Physics Chapter 8- Rotational of a Rigid Body

DF015 CHAPTER 8

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CHAPTER 8: Rotation of a rigid body

(8 Hours)

DF025 CHAPTER 8

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At the end of this chapter, students should be able to: a) Define and use:

angular displacement () average angular velocity (av)

instantaneous angular velocity () average angular acceleration (av)

instantaneous angular acceleration ().

Learning Outcome:

8.1 Rotational kinematics (2 hours)

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b) Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use;

c) Use equations for rotational motion with constant angular acceleration;

, , .

r

vrararvrθs ct

22 ; ; ;

αθωω 20

2 2αtωω 0 20 ttωθ

2

1

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8.1 Rotational kinematics a)(i) Angular displacement, is defined as an angle through which a point or line has

been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad). Figure 7.1 shows a point P on a rotating compact disc (CD)

moves through an arc length s on a circular path of radius r about a fixed axis through point O.

Figure 7.1

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From Figure 7.1, thus

Others unit for angular displacement is degree () and revolution (rev). Conversion factor :

Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise. Negative – if the rotational motion is clockwise.

360rad 2πrev 1

r

sθ OR rθs

where radianin nt)displaceme(angular angle :θlength arc :s

circle theof radius :r

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Average angular velocity, av

is defined as the rate of change of angular displacement. Equation :

Instantaneous angular velocity, is defined as the instantaneous rate of change of angular

displacement. Equation :

a)(ii)(iii) Angular velocity

t

θ

tt

θθω

12

12

av

dt

dθ

t

θ0t

limit

where radianin nt displacemeangular final :2θ

interval time :tradianin nt displacemeangular initial :1θ

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It is a vector quantity. The unit of angular velocity is radian per second (rad s-1) Others unit is revolution per minute (rev min1 or rpm)

Conversion factor:

Note : Every part of a rotating rigid body has the same angular

velocity.

Direction of the angular velocity Its direction can be determine by using right hand grip rule

where

11 s rad 30

s rad 60

2 rpm 1

Thumb : direction of angular velocity

Curl fingers : direction of rotation

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Figures 7.2 and 7.3 show the right hand grip rule for determining the direction of the angular velocity.

Figure 7.2

Figure 7.3

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The angular displacement, of the wheel is given by

where in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine

a. the angle, in degree, at time 2.2 s and 4.8 s,

b. the distance that a particle on the rim moves during that time

interval,

c. the average angular velocity, in rad s1 and in rev min1 (rpm),

between 2.2 s and 4.8 s,

d. the instantaneous angular velocity at time 3.0 s.

Example 1 :

ttθ 2 5

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Solution :

a. At time, t1 =2.2 s :

At time, t2 =4.8 s :

m 0.282

0.56

2

dr

2.22.25 2 1θ

rad 221θ

1261rad π

180 rad 22

1θ

4.84.85 2 2θ

rad 1102θ

6303rad π

180 rad 110

2θ

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Solution :

b. By applying the equation of arc length,

Therefore

c. The average angular velocity in rad s1 is given by

m 0.282

0.56

2

dr

rθs

m 24.6s

12rrs 221100.28 s

12

12

ttt

θω

av

1av s rad 33.9 ω

2.24.8

22110av

ω

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Solution :

c. and the average angular velocity in rev min1 is

d. The instantaneous angular velocity as a function of time is

At time, t =3.0 s :

1av min rev 324 ω

min 1

s 60

rad 2

rev 1

s 1

rad 33.9avω

OR rpm 324

ttdt

dω 25

dt

dθω

110 tω

13.010 ω1s rad 29 ω

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A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the diver’s average angular (rotational) velocity during a dive.

(Given g = 9.81 m s2)

Solution :

Example 2 :

0yu00θ

m 10

waterrev 2.51θ

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Solution :

From the diagram,

Thus

Therefore the diver’s average angular velocity is s 1.43t

rad 5ππ22.5 1θm 10ys

2yy gttus

2

1

29.812

1010 t

t

θθω 01

av

1av s rad 11.0 ω

1.43

05πav

ω

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Average angular acceleration, av

is defined as the rate of change of angular velocity. Equation :

Instantaneous angular acceleration, is defined as the instantaneous rate of change of angular

velocity. Equation :

a)(iv)(v) Angular acceleration

t

ω

tt

ωω

12

12av

dt

dω

t

ωα

t

0

limit

where locityangular ve final :2ω

interval time :tlocityangular ve initial :1ω

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Figure 7.4

It is a vector quantity. The unit of angular acceleration is rad s2. Note:

If the angular acceleration, is positive, then the angular

velocity, is increasing. If the angular acceleration, is negative, then the angular

velocity, is decreasing.

Direction of the angular acceleration If the rotation is speeding up, and in the same direction as

shown in Figure 7.4.

α

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Figure 7.5

If the rotation is slowing down, and have the opposite direction as shown in Figure 7.5.

Example 3 :

The instantaneous angular velocity, of the flywheel is given

by

where in radian per second and t in seconds.

Determine

a. the average angular acceleration between 2.2 s and 4.8 s,

b. the instantaneous angular acceleration at time, 3.0 s.

α

23 tt8ω

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Solution :

a. At time, t1 =2.2 s :

At time, t2 =4.8 s :

Therefore the average angular acceleration is

2.22.28 23 1ω1s rad 80.3 1ω

2av s rad 301 α

4.84.88 23 2ω1s rad 862 2ω

12

12

tt

ωωα

av

2.24.8

80.3862av

α

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Solution :

b. The instantaneous angular acceleration as a function of time is

At time, t =3.0 s :

238 ttdt

dα

dt

dωα

ttα 224 2

3.023.024 2 α2s rad 210 α

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Exercise 8.1a :1. If a disc 30 cm in diameter rolls 65 m along a straight line

without slipping, calculate

a. the number of revolutions would it makes in the process,

b. the angular displacement would be through by a speck of

gum on its rim.

ANS. : 69 rev; 138 rad

2. During a certain period of time, the angular displacement of a swinging door is described by

where is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration

a. at time, t =0,

b. at time, t =3.00 s.

ANS. : 5.00 rad, 10.0 rad s1, 4.00 rad s2; 53.0 rad, 22.0 rad s1, 4.00 rad s2

22.0010.05.00 ttθ

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At the end of this chapter, students should be able to: Relate parameters in rotational motion with their

corresponding quantities in linear motion. Write and use;

Learning Outcome:8.1.b) Relationship between linear and rotational motion (½ hour)

r

vrararvrθs ct

22 ; ; ;

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8.1.b) Relationship between linear and rotational motion Relationship between linear velocity, v and

angular velocity, When a rigid body is rotates about rotation axis O , every

particle in the body moves in a circle as shown in the Figure 7.6.

v

s

y

xr

P

O

Figure 8.6

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Point P moves in a circle of radius r with the tangential velocity

v where its magnitude is given by

The direction of the linear (tangential) velocity always tangent to the circular path.

Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle, r is changing depend on the position of the particle.

dt

dsv

dt

drv

rθs

rv

and

Simulation 8.1

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ta

caa

x

y

P

O

If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 7.7.

Relationship between tangential acceleration, at and

angular acceleration,

Figure 8.7

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The components are tangential acceleration, at and

centripetal acceleration, ac given by

but

The vector sum of centripetal and tangential acceleration of

a particle in a rotating body is resultant (linear) acceleration, a given by

and its magnitude,

dt

dvat

dt

drat

rat

rωv and

vrr

va 2

2

c

ct aaa

2c

2t aaa

Vector form

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At the end of this chapter, students should be able to: Write and use equations for rotational motion with

constant angular acceleration;

Learning Outcome:

8.1.c) Rotational motion with uniform angular acceleration (1/2 hour)

αtωω 0

αθωω 20

2 2

20 ttωθ

2

1

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8.1.c) Rotational motion with uniform angular acceleration Table 8.1 shows the symbols used in linear and rotational

kinematics.

Table 8.1

Linear motion Quantity

Rotational motion

s θDisplacement

u 0ωInitial velocity

v ωFinal velocity

a αAcceleration

t tTime

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Table 8.2 shows the comparison of linear and rotational motion with constant acceleration.

Linear motion Rotational motion

constanta

atuv

constantα

αtωω 0

2atuts2

1 2

0 αttωθ2

1

asuv 22 2 αθωω 20

2 2

tuvs 2

1 tωωθ 02

1

where in radian. Table 8.2

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A car is travelling with a velocity of 17.0 m s1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s2 for 5.00 s, calculate

a. the number of revolutions of the wheels during this period,

b. the angular speed of the wheels after 5.00 s.

Solution :

a. The initial angular velocity is

and the angular acceleration of the wheels is given by

Example 4 :

s 5.00 ,s m 2.00 ,m 0.48 ,s m 17.0 21 taru

0rωu 0ω0.4817.0 1s rad 35.4 0ω

α0.482.00 rαa

2s rad 4.17 α

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Solution :

a. By applying the equation of rotational motion with constant

angular acceleration, thus

therefore

b. The angular speed of the wheels after 5.00 s is

2

2

1αttωθ 0

rad 229θ

s 5.00 ,s m 2.00 ,m 0.48 ,s m 17.0 21 taru

25.004.172

15.0035.4 θ

rev 36.5rad 2π

rev 1 rad 229

θ

αtωω 0

1s rad 56.3 ω

5.004.1735.4 ω

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The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. The wheels have a diameter of 70 cm.

a. Calculate the angular acceleration.

b. If the bicycle continues to decelerate at this rate, determine the

time taken for the bicycle to stop.

Solution :

Example 5 :

,m 0.35 2

0.70 ,rad 60π2π30 rθ

,s m 13.9s 3600

h 1

km 1

m10

h 1

km 50.0 13

u

13

s m 9.72s 3600

h 1

km 1

m10

h 1

km 35.0

v

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Solution :

a. The initial angular speed of the wheels is

and the final angular speed of the wheels is

therefore

b. The car stops thus

Hence

0rωu 0ω0.3513.9 1s rad 39.7 0ω

rωv ω0.359.72 1s rad 27.8 ω

αθωω 0 222

2s rad 2.13 α 60π239.727.8 22 α

0ω 1s rad 27.8 0ωand

αtωω 0

s 13.1t

t2.1327.80

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A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s-1. The angular acceleration of the blade is 0.750 rev s-2. Determine

a. the angular velocity after 4.00 s,

b. the number of revolutions for the blade turns in this time interval,

c. the tangential speed of a point on the tip of the blade at time,

t =4.00 s,

d. the magnitude of the resultant acceleration of a point on the tip

of the blade at t =4.00 s.

Solution :

a. Given t =4.00 s, thus

Example 6 :

,s rad 0.300π2π0.150 ,m 0.400 1 0ωr2s rad 1.50π2π0.750 α

αtωω 0 1s rad 19.8 ω

4.001.50π0.300π ω

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Solution :

b. The number of revolutions of the blade is

c. The tangential speed of a point is given by

2

2

1αttωθ 0

rad 41.5θ

24.001.502

14.000.300 θ

rev 6.61rad 2π

rev 1 rad 41.5

θ

rωv 19.80.400v

1s m 7.92 v

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Solution :

d. The magnitude of the resultant acceleration is

22tc aaa

2s m 157 a

2

22

rαr

va

2

22

1.50π0.4000.400

7.92

a

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Calculate the angular velocity of

a. the second-hand,

b. the minute-hand and

c. the hour-hand,

of a clock. State in rad s-1.

d. What is the angular acceleration in each case?

Solution :

a. The period of second-hand of the clock is T = 60 s, hence

Example 7 :

Tω

2π

1s rad 0.11 ω

60

2πω

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Solution :

b. The period of minute-hand of the clock is T = 60 min = 3600 s,

hence

c. The period of hour-hand of the clock is T = 12 h = 4.32 104 s,

hence

d. The angular acceleration in each cases is zero.

3600

2πω

13 s rad 101.74 ω

4104.32

2π

ω

14 s rad 101.45 ω

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A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s2, calculate the distance travelled by the coin before coming to rest.

Solution :

The radius of the coin is

Example 8 :

m 102.40 2d

1s rad 18 0ω

s

2s rad 1.90 α

1s rad 0 ω

m 101.202

2d

r

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Solution :

The initial speed of the point at the edge the coin is

and the final speed is

The linear acceleration of the point at the edge the coin is given by

Therefore the distance travelled by the coin is

0rωu 18101.20 2u

1s m 0.216 u1s m 0 v

rαa 1.90101.20 2 a

22 s m 102.28 a

asuv 222 s22 102.2820.2160

m 1.02s

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Exercise 8.1b&c :1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev

min-1 about its central axis. Determine

a. its angular speed,

b. the tangential speed at a point 3.00 cm from its centre,

c. the radial acceleration of a point on the rim,

d. the total distance a point on the rim moves in 2.00 s.

ANS. : 126 rad s1; 3.77 m s1; 1.26 103 m s2; 20.1 m

2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculate

a. its angular velocity in rad s1,

b. the linear speed and the radial acceleration of a point on the

edge of the grinding wheel.

ANS. : 262 rad s1; 46 m s1, 1.2 104 m s2

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Exercise 8.1b&c :3. A rotating wheel required 3.00 s to rotate through 37.0

revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel.

ANS. : 13.6 rad s2

4. A wheel rotates with a constant angular acceleration of 3.50 rad s2.

a. If the angular speed of the wheel is 2.00 rad s1 at t =0,

through what angular displacement does the wheel rotate in

2.00 s.

b. Through how many revolutions has the wheel turned during

this time interval?

c. What is the angular speed of the wheel at t = 2.00 s?

ANS. : 11.0 rad; 1.75 rev; 9.00 rad s1

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Exercise 8.1b&c :5. A bicycle wheel is being tested at a repair shop. The angular

velocity of the wheel is 4.00 rad s-1 at time t = 0 , and its angular acceleration is constant and equal 1.20 rad s-2. A spoke OP on the wheel coincides with the +x-axis at time t = 0 as shown in Figure 7.8.

a. What is the wheel’s angular velocity at t = 3.00 s?

b. What angle in degree does the spoke OP make with the

positive x-axis at this time?

ANS. : 0.40 rad s1; 18

Figure 8.8

x

y

PO

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At the end of this chapter, students should be able to:

a) Define and use torque,

b) State and use conditions for equilibrium of rigid body:

Examples of problems :

Fireman ladder leaning on a wall, see-saw, pivoted / suspended horizontal bar.

Sign convention for moment or torque :

+ve : anticlockwise

ve : clockwise

Learning Outcome:8.2 Equilibrium of a uniform rigid body (2 hours)

τ

0 , 0 , 0 τFF yx

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8.2 Equilibrium of a rigid bodyNon-concurrent forces

is defined as the forces whose lines of action do not pass through a single common point.

The forces cause the rotational motion on the body. The combination of concurrent and non-concurrent forces cause

rolling motion on the body. (translational and rotational motion)

Figure 5.11 shows an example of non-concurrent forces.

2F

3F

1F

Figure 8.2

4F

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Torque (moment of a force), The magnitude of the torque is defined as the product of a

force and its perpendicular distance from the line of action of the force to the point (rotation axis).

OR

Because of

where r : distance between the pivot point (rotation axis) and the point of application of force.

Thus

Fdτ

force theof magnitude :Farm)(moment distancelar perpendicu : d

torque theof magnitude : τwhere

sinrd

sin FrrF

and between angle : where

OR Fr

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It is a vector quantity. The dimension of torque is

The unit of torque is N m (newton metre), a vector product unlike the joule (unit of work), also equal to a newton metre, which is scalar product.

Torque is occurred because of turning (twisting) effects of the forces on a body.

Sign convention of torque: Positive - turning tendency of the force is anticlockwise. Negative - turning tendency of the force is clockwise.

The value of torque depends on the rotation axis and the magnitude of applied force.

22 TMLdF

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Case 1 : Consider a force is applied to a metre rule which is pivoted at

one end as shown in Figures 5.12a and 5.12b.

Figure 8.12a

F

F

θ

Figure 8.12b

Pivot point (rotation axis)

Fdτ

θrd sin

θFrFdτ sin

(anticlockwise)

(anticlockwise)r

Point of action of a force

Line of action of a force

d

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O

Figure 8.13

2θ

Case 2 : Consider three forces are applied to the metre rule which is

pivoted at one end (point O) as shown in Figures 5.13.

Caution : If the line of action of a force is through the rotation axis

then

1F

1θ

111 θrd sin

321 ττττ O

Therefore the resultant (nett) torque is

3F

2F

1r

0sin 333333 θrFdFτ

222 θrd sin

111111 θrFdFτ sin

222222 θrFdFτ sin

2r

2211 dFdFτ O

θFrτ sin0τ

and 0θSimulation 8.2

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Determine a resultant torque of all the forces about rotation axis, O in the following problems.

a.

Example 4 :

m 5

N 102F

m 5 N 301F

m 3

m 3

N 203F

m 10

m 6O

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b.

Example 4 :

m 5

N 102F

m 5

N 301F

m 3

m 3

N 254F

N 203F

m 10

α

m 6O β

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m 5m 5

m 10

m 6O

Solution :

a.

Force Torque (N m), o=Fd=Frsin

1F

90330

2F

50510

N 102FN 301F

N 203F

m 31d

m 52d

3F

0The resultant torque:

m N 405090 Oτ(clockwise)

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m 5

m 10

m 3

m 6

m 5

Solution :

b.

Force Torque (N m), o=Fd=Frsin

1F

90330

2F

51.50.515520sin βrF33F

0 The resultant torque:51.590 Oτ

(clockwise)

N 102F

N 301F

0.51553

3sin

22

β

O

N 203F

N 254Fα

β

m 31dβ

m 5r

4F

0

3d

m N 38.5 Oτ

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8.2 Equilibrium of a rigid body Rigid body is defined as a body with definite shape that

doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it.

If the rigid body is in equilibrium, means the body is translational and rotational equilibrium.

There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must

be zero.

0nettFF

OR

0 , 0 , 0 zyx FFF

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The vector sum of all external torques acting on a rigid body must be zero about any rotation axis.

This ensures rotational equilibrium. This is equivalent to the three independent scalar

equations along the direction of the coordinate axes,

Centre of gravity, CG is defined as the point at which the whole weight of a body

may be considered to act. A force that exerts on the centre of gravity of an object will

cause a translational motion.

0nettτ

0 , 0 , 0 zyx τττ

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Figures 5.14 and 5.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere

rod – refer to the midway point between its end.

sphere – refer to geometric centre.

2

l

2

l

CG

CGl

Figure 5.14

Figure 5.15

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56

5.3.4 Problem solving strategies for equilibrium of a rigid body

The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagram of the system to help

conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and

construct a table to resolve the forces into their components and to determine the torque by each force.

Apply the condition for equilibrium of a rigid body :

Solve the equations for the unknowns.

0xF 0yF; and 0τ

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A hanging flower basket having weight, W2 =23 N is hung out over

the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is

counterbalanced by a body of weight, W1 as shown in Figure 5.16.

If the mass of the beam is 3.0 kg, calculate

a. the weight, W1 needed,

b. the force exerted on the beam at point O.

(Given g =9.81 m s2)

Example 5 :

1W2W

A BO35 cm

75 cm

Figure 5.16

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Solution :

The free body diagram of the beam :

Let point O as the rotation axis.

N 23 ;kg 3 2Wm

0.75 mA B

OCG

1W

2W

N

gm

0.35 m

0.55 m 0.55 m

0.20 m

Force y-comp. (N) Torque (N m), o=Fd=Frsin

1W

1W

gm 9.813 5.880.2029.4

11 WW 0.750.75

2W

23 8.050.3523

N

N 029.4

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Solution :

Since the beam remains at rest thus the system in equilibrium.

a. Hence

b.

N 2.891W

0 yFand

0Oτ

05.888.050.75 1W

N 55.3N

029.423 NW1

029.4232.89 N

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A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 5.17. The height of the end A of the ladder is 8.0 m from the rough floor.

a. Determine the horizontal and vertical

forces the floor exerts on the end B of

the ladder when a firefighter of mass

60 kg is 3.0 m from B.

b. If the ladder is just on the verge of

slipping when the firefighter is 7.0 m

up the ladder , Calculate the coefficient

of static friction between ladder and

floor.


Example 6 :

A

B

smooth wall

rough floor

Figure 5.17

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Solution :

a. The free body diagram of the ladder :

Let point B as the rotation axis.

kg 60 ;kg 5.0 fl mm

A

B

CG

gm f

1N

gml

2N

α

m 8.0m 10

m 3.0

m 5.0

Force x-comp. (N)

y-comp. (N)

Torque (N m), B=Fd=Frsin

gml

1N1N

0.810

8sin α

sf

gm f

49.1 0.6

10

6sin β

2N

sf

0

5890

2N

0

0

m 6.0

αβ βsin5.049.1

β

β

147

0

βsin3.05891060

αN1 sin101N8

0

0 sf

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Solution :

Since the ladder in equilibrium thus

0 Bτ

081060147 1N

N 1511N

0 xF

0 s1 fNN 151sfHorizontal force:

0 yF

058949.1 2NN 6382NVertical force:

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m 10

A

B

m 8.0

m 6.0

m 5.0 α

β

Solution :

b. The free body diagram of the ladder :

Let point B as the rotation axis.

0.6sin 0.8;sin βα

gm f

gml

sf

m 7.0

Force x-comp. (N)

y-comp. (N)

Torque (N m), B=Fd=Frsin

gml

1N1N

2s Nμ

gm f

49.1

2N

sf

0

5890

2N

0

0

α

βsin5.049.1

β

β147

0

βsin7.05892474

αN1 sin101N8

0

0

2N

1N

DF025 CHAPTER 8

64

Solution :

Consider the ladder stills in equilibrium thus

0 Bτ

082474147 1N

N 3281N

0 xF

0 2s1 NμN

0 yF

058949.1 2NN 6382N

0638328 sμ0.514sμ

DF025 CHAPTER 8

65

Figure 5.18

A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 5.18. A cable at an angle 30 with the beam helps to support the light.

a. Sketch a free body diagram of the beam.

b. Determine

i. the tension in the cable,

ii. the force exerted on the beam by the

pole.


Example 7 :

DF025 CHAPTER 8

66

Solution :

a. The free body diagram of the beam :

b. Let point O as the rotation axis.

kg 10.0 ;kg 20.0 bf mm

Force x-comp. (N) y-comp. (N) Torque (N m), o=Fd=Frsingm f

l1960 196

O CG

gm f

gmb

T

S

30

l

l0.5

gmb

ll 49.10.598.1 0 98.1

T

TlTl 0.530sin 30cosT 30sinT

S

xS yS 0

DF025 CHAPTER 8

67

Solution :

b. The floodlight and beam remain at rest thus

i.

ii.

0 Oτ

00.549.1196 TlllN 490T

0 xF

0cos xS30T

N 424xS

0 yF

030sin98.1196 yST

N 49.1yS

DF025 CHAPTER 8

68

Solution :

b. ii. Therefore the magnitude of the force is

and its direction is given byN 427S

2y

2x SSS

22S 49.1424

x

y

S

Sθ 1tan

6.61θ

424

49.1tan 1θ

from the +x-axis anticlockwise

DF025 CHAPTER 8

69

Exercise 8.2 :

Use gravitational acceleration, g = 9.81 m s2

1.

Figure 5.19 shows the forces, F1 =10 N, F2= 50 N and F3=

60 N are applied to a rectangle with side lengths, a = 4.0 cm

and b = 5.0 cm. The angle is 30. Calculate the resultant torque about point D.

ANS. : -3.7 N m

D

AB

C γ

1F

3F

2F

Figure 8.19

a

b

DF025 CHAPTER 8

70

Figure 5.20

Exercise 8.2 :2.

A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 5.20. The fulcrum is under the centre of gravity of the board. Determine

a. the magnitude of the force exerted by the fulcrum on the

board,

b. where the father should sit from the fulcrum to balance the

system.

ANS. : 1128 N; 1.31 m

DF025 CHAPTER 8

71

3.

A traffic light hangs from a structure as show in Figure 5.21. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine

a. the tension in the horizontal massless cable CD,

b. the vertical and horizontal components of the force exerted

by the pivot A on the aluminum pole.

ANS. : 248 N; 197 N, 248 N

Figure 5.21

Exercise 8.2 :

DF025 CHAPTER 8

72

4.

A uniform 10.0 N picture frame is supported by two light string

as shown in Figure 5.22. The horizontal force, F is applied for holding the frame in the position shown.

a. Sketch the free body diagram of the picture frame.

b. Calculate

i. the tension in the ropes,

ii. the magnitude of the horizontal force, F .

ANS. : 1.42 N, 11.2 N; 7.20 N

Exercise 8.2 :

Figure 5.22

F

50.0

cm 15.0

cm 30.0

DF025 CHAPTER 8

73

8.3 Rotational Dynamics (1 hour)

Centre of mass (CM) is defined as the point at which the whole mass of a body

may be considered to be concentrated.

Its coordinate (xCM, yCM) is given the expression below:

n

1ii

n

1iii

CM

m

xm

x

n

1ii

n

1iii

CM ;

m

ym

y

where particle i theof mass : thim

particle i theof coordinate : thi xx

particle i theof coordinate : thi yy

DF025 CHAPTER 8

74

Two masses, 2 kg and 4 kg are located on the x-axis at x =2 m

and x =5 m respectively. Determine the centre of mass of this system.

Solution :

Example 9 :

0 m 2x m 5

kg 4 kg; 2 21 mm1m 2m

21

22112

1ii

2

1iii

CM mm

xmxm

m

xm

x

42

5422CM

x

0 from m 4CM xx

CM

m 4

2 m from m1

OR

DF025 CHAPTER 8

75

A system consists of three particles have the following masses and coordinates :

(1) 2 kg, (1,1) ; (2) 4 kg, (2,0) and (3) 6 kg, (2,2).

Determine the coordinate of the centre of mass of the system.

Solution :

The x coordinate of the CM is

Example 10 :

kg 3 kg; 2 kg; 1 321 mmm

321

3322113

1ii

3

1iii

CM mmm

xmxmxm

m

xm

x

642

262412CM

x 1.83CM x

DF025 CHAPTER 8

76

Solution :

The y coordinate of the CM is

Therefore the coordinate of the CM is

321

3322113

1ii

3

1iii

CM mmm

ymymym

y

ym

y

642

260412CM

y 1.17CM y

1.17,1.83

DF025 CHAPTER 8

77

Figure 7.9 shows a rigid body about a fixed axis O with angular

velocity .

is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis.

Moment of inertia, I

1m

2mnm

3m

1r2r

3rnr

O

Figure 7.9

DF025 CHAPTER 8

78

OR

It is a scalar quantity. Moment of inertia, I in the rotational kinematics is analogous

to the mass, m in linear kinematics. The dimension of the moment of inertia is M L2. The S.I. unit of moment of inertia is kg m2. The factors which affect the moment of inertia, I of a rigid body:

a. the mass of the body,b. the shape of the body,c. the position of the rotation axis.

n

1i

2ii

2nn

233

222

211 ... rmrmrmrmrmI

axisrotation about body rigid a of inertia ofmoment : Iparticle of mass : m

axisrotation the toparticle thefrom distance : r

where

DF025 CHAPTER 8

79

Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects

about axes through the centre of mass.

Shape Diagram Equation

2CM MRI

2CM 2

1MRI

Hoop or ring or thin cylindrical

shell

Solid cylinder or disk

CM

CM

DF025 CHAPTER 8

80

CM




2CM 12

1MLI

Uniform rod or long thin rod with

rotation axis through the

centre of mass.

CM

2CM 5

2MRI Solid Sphere

DF025 CHAPTER 8

81




2CM 3

2MRI Hollow Sphere or

thin spherical shell

CM

Table 7.3

DF025 CHAPTER 8

82

Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 7.10.

The spheres are connected by light rods . Determine the moment of inertia of the system about an axis

a. through point O,

b. along the line AB.

Example 11 :

cm 250

cm 60

cm 60

kg 2 kg 3

kg 4kg 5

OA B

Figure 7.10

DF025 CHAPTER 8

83

Solution :

a. rotation axis about point O,

Since r1= r2= r3= r4= r thus

and the connecting rods are light therefore

kg 5 kg; 4 kg; 3 kg; 2 4321 mmmm

1r m 0.6

1m 2m

3m4m

O

2r

4r 3r

m 1.25

m 1.391.250.6 22 r

244

233

222

211O rmrmrmrmI 54321.39 2

43212

O mmmmrI2

O m kg 27.0I

DF025 CHAPTER 8

84

Solution :

b. rotation axis along the line AB,

r1= r2= r3= r4= r=0.6 m therefore

kg 5 kg; 4 kg; 3 kg; 2 4321 mmmm

244

233

222

211AB rmrmrmrmI

54320.6 2AB I

2AB m kg 5.04I

1m 2m

3m4m

A B1r 2r

4r 3r

43212

AB mmmmrI

DF025 CHAPTER 8

85

Parallel-Axis Theorem (Steiner’s Theorem) States that the moment of inertia, I about any axis parallel to

and a distance, d away from the axis through the centre of

mass, ICM is given by

2CM MdII

axisrotation new aabout inertia ofmoment :I

body rigid theof mass : M

CM through axisan about inertia ofmoment :CMI

axis original theand axis new ebetween th distance : d

where

DF025 CHAPTER 8

86

Determine the moment of inertia of a solid cylinder of radius R and

mass M about an axis tangent to its edge and parallel to its symmetry axis as shown in Figure 7.11.

Given the moment of inertia of the solid cylinder about axis through the centre of mass is

Example 12 :

Figure 7.11

CM

d

2CM 2

1MRI

DF025 CHAPTER 8

87

CM

Solution :

From the diagram, d = RBy using the parallel axis theorem,

CM

Initiald

Final

2CM MdII

22

2

1MRMRI

2

2

3MRI

2CM 2

1MRI

DF025 CHAPTER 8

88

Relationship between torque, and angular acceleration, Consider a force, F acts on a rigid body freely pivoted on an

axis through point O as shown in Figure 7.12.

The body rotates in the anticlockwise direction and a nett torque is produced.

Torque,

1m

2m

nm1r

2r

nr

O

1a

na

2a

F

Figure 7.12

DF025 CHAPTER 8

89

A particle of mass, m1 of distance r1 from the rotation axis O will

experience a nett force F1 . The nett force on this particle is

The torque on the mass m1 is

The total (nett) torque on the rigid body is given by

111 amF

αrmF 111

αra 11 and

90sin111 Fr 2

111 rm

n

iiirm

1

2

2222

211 ... nnrmrmrm

I

Irmn

iii

1

2 and

DF025 CHAPTER 8

90

From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration.

Note :

I, eNett torqu

maFforce,Nett

is analogous to the

DF025 CHAPTER 8

91

Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 7.13.

Calculate,

a. the nett torque on the disc.

b. the magnitude of angular acceleration influence by the disc.

( Use the moment of inertia, )

Example 13 :

Figure 7.13

2CM 2

1MRI

1F

O

cm 30.0

2F

DF025 CHAPTER 8

92

Solution :

a. The nett torque on the disc is

b. By applying the relationship between torque and angular

acceleration,

kg 5.00 ;m 0.30 MR

21 2121 FFRRFRF

3.1060.530.0 m N 1.41

2

2

1MR I

2s rad 6.27

230.000.5

2

141.1

DF025 CHAPTER 8

93

A wheel of radius 0.20 m is mounted on a frictionless horizontal

axis. The moment of inertia of the wheel about the axis is

0.050 kg m2. A light string wrapped around the wheel is attached

to a 2.0 kg block that slides on a horizontal frictionless surface. A

horizontal force of magnitude P = 3.0 N is applied to the block as

shown in Figure 7.14. Assume the string does not slip on the

wheel.

a. Sketch a free body diagram of the wheel and the block.

b. Calculate the magnitude of the angular acceleration of the wheel.

Example 14 :

Figure 7.14

DF025 CHAPTER 8

94

Solution :

a. Free body diagram :

for wheel,

for block,

kg 2.0 N; 3.0 ;m kg 0.050 ;m 0.20 2 mPIR

W

T

S

N

T

bW

P

a

DF025 CHAPTER 8

95

Solution :

b. For wheel,

For block,

By substituting eq. (1) into eq. (2), thus

Iατ

IαRT R

IαT (1)

maF maTP (2)

kg 2.0 N; 3.0 ;m kg 0.050 ;m 0.20 2 mPIR

Rαa maR

IαP

and

mRαR

IαP

αα0.202.0

0.20

0.0503.0

2s rad 4.62 α

DF025 CHAPTER 8

96

An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 7.15. The pulley has a moment of inertia 0.020 kg m2

about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determine

a. the linear acceleration of the object,

b. the angular acceleration of the pulley,

c. the tension in the string,

d. the liner velocity of the object,

e. the distance travelled by the object. (Given g = 9.81 m s-2)

Example 15 :

Figure 7.15

R

kg 1.50

DF025 CHAPTER 8

97

Solution :

a. Free body diagram :

for pulley,

for block,

W

a

T

S Iατ

IαRT R

aα and

R

aIRT

2R

IaT (1)

T

gm

maF

maTmg (2)

DF025 CHAPTER 8

98

Solution :

a. By substituting eq. (1) into eq. (2), thus

b. By using the relationship between a and , hence

2s m 7.36 a

maR

Iamg

2

kg; 1.50 ;m kg 0.020 ;m 0.20 2 mIRs 0.3 0; tu

Rαa

aa

1.500.20

0.0209.811.50

2

α0.207.36 2s rad 36.8 α

DF025 CHAPTER 8

99

Solution :

c. From eq. (1), thus

d. By applying the equation of liner motion, thus

e. The distance travelled by the object in 0.3 s is

N 3.68T

kg; 1.50 ;m kg 0.020 ;m 0.20 2 mIRs 0.3 0; tu

atuv 0.37.360 v

m 0.331s

2R

IaT

20.20

7.360.020T

1s m 2.21 v(downwards)

2at2

1uts

20.37.3602

1s

DF025 CHAPTER 8

100

Exercise 8.3 :Use gravitational acceleration, g = 9.81 m s2

1. Three odd-shaped blocks of chocolate have following masses and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m);

0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m).

Determine the coordinates of the centre of mass of the system of three chocolate blocks.

ANS. : (0.044 m, 0.056 m)

2. Figure 7.16 shows four masses that are held at

the corners of a square by a very light

frame. Calculate the moment of inertia

of the system about an axis perpendicular

to the plane

a. through point A, and

b. through point B.

ANS. : 0.141 kg m2; 0.211 kg m2

cm 80

cm 80

g 150 g 150

g 07

g 70

cm 40

A

B

Figure 7.16

DF025 CHAPTER 8

101

2s m 2.00

2T

1T

Exercise 8.3 :3. A 5.00 kg object placed on a

frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 7.17. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s2.a. Sketch free body diagrams of both objects and pulley.

b. Calculate T1 and T2 the tensions

in the string.c. Determine I.

ANS. : 10.0 N, 70.3 N; 1.88 kg m2

Figure 7.17

DF025 CHAPTER 8

102

At the end of this chapter, students should be able to: a) Solve problems related to:

Rotational kinetic energy,

work,

power,

Learning Outcome:

8.4 Work and Energy of Rotational Motion(2 hours)

2

2

1IωK r

τωP

τθW

DF025 CHAPTER 8

103

Rotational kinetic energy and powerRotational kinetic energy, Kr

Consider a rigid body rotating about the axis OZ as shown in Figure 7.18.

Every particle in the body is in the circular motion about point O.

1m

2m

nm

3m

1r2r

3r

nr

O

1v

2v

3vnv

Z

Figure 7.18

DF025 CHAPTER 8

104

The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by

2nn

233

222

211r vmvmvmvmK

2

1...

2

1

2

1

2

1

22nn

2233

2222

2211r ωrmωrmωrmωrmK

2

1...

2

1

2

1

2

1

2nn

233

222

211

2r rmrmrmrmωK ...

2

1

2r IωK

2

1

n

1i

2ii

2r rmωK

2

1Irm

n

1i

2ii

and

DF025 CHAPTER 8

105

From the formula for translational kinetic energy, Ktr

After comparing both equations thus

For rolling body without slipping, the total kinetic energy of

the body, K is given by

2tr mv

2

1K

is analogous to vI is analogous to m

rtr KKK

energy kinetic onal translati: trK

energy kinetic rotational : rK

where

DF025 CHAPTER 8

106

A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25 to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate

a. the total kinetic energy of the sphere,

b. the linear speed of the sphere,

c. the angular speed about the centre of mass.

(Given the moment of inertia of solid sphere is and

the gravitational acceleration, g = 9.81 m s2)

Example 16 :

2mRI5

2CM

DF025 CHAPTER 8

107

Solution :

a. From the principle of conservation of energy,

kg 10.0 ;m 0.15 mR

fi EE

Kmgh 25sinmgsK

J 31.1K

25sin0.759.8110.0K

m 0.75s

25sinsh

v CM 25

R

DF025 CHAPTER 8

108

Solution :

b. The linear speed of the sphere is given by

c. By using the relationship between v and , thus

kg 10.0 ;m 0.15 mR

rtr KKK 22 IωmvK2

1

2

1

R

vω

1s m 2.11 v

and

222

R

vmRmvK

5

2

2

1

2

1

2mvK10

7

1s rad 14.1 ω

2v10.010

731.1

Rωv ω0.152.11

DF025 CHAPTER 8

109

The pulley in the Figure 7.19 has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim.

Calculate the speed of the 5.00 kg block just before it strikes the floor.

(Given g = 9.81 m s2)

Example 17 :

kg 2.00

kg 5.00

m 7.00

Figure 7.19

DF025 CHAPTER 8

110

Solution :

The moment of inertia of the pulley,m 7.00 m; 0.120 kg; 2.00 ;kg 5.00 21 hRmm

292

2432 m kg 105.5

cm 1

m10

g 1

kg 10cm 1g 0.055

I

2m

1m

m 7.00

Initial

2m

1mm 7.00

v

v

Final

1i UE 2r2tr1trf UKKKE

DF025 CHAPTER 8

111

Solution :

By using the principle of conservation of energy, thus

fi EE

2r2tr1tr1 UKKKU

ghmIωvmvmghm 222

22

11 2

1

2

1

2

1

2

212

21 2

1

2

1

R

vImmvghmm

2

92

0.120105.5

2

12.005.00

2

17.009.812.005.00

v

v

1s m 7.67 v

m; 0.120 kg; 2.00 ;kg 5.00 Rmm 2129 m kg 105.5 m; 7.00 Ih

DF025 CHAPTER 8

112

Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 7.20.

The work done by the tangential force is given by

Work, W

Figure 7.20

F

ds

O

dR

R

FdsdW FRdθdW

2

1

θ

θτdθ dW

2

1

θ

θτdθ W

Rdθds and

DF025 CHAPTER 8

113

If the torque is constant thus

Work-rotational kinetic energy theorem states

12W 2

1

dW

W

torque: τntdisplacemeangular in change : Δθ

where

done work : W

irfrr KKKW

20

2

2

1

2

1IωIωW

is analogous to the FsW

DF025 CHAPTER 8

114

From the definition of instantaneous power,

Caution : The unit of kinetic energy, work and power in the

rotational kinematics is same as their unit in translational kinematics.

Power, P

dt

dWP τdθdW and

dt

τdθP

τω P

ωdt

dθand

is analogous to the FvP

DF025 CHAPTER 8

115

A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determine

a. the rotational kinetic energy of the merry-go-round,

b. the work done by the force on the merry-go-round,

c. the average power supplied by the force.

(Given g = 9.81 m s2)

Solution :

Example 18 :

F

m 2.40R

DF025 CHAPTER 8

116

Solution :

a. By applying the relationship between nett torque and angular

acceleration, thus

Use the equation of rotational motion with uniform angular

acceleration,

Therefore the rotational kinetic energy for 15.0 s is

Iατ IαRF α210018.02.40

22 s rad 102.06 α

αtωω 0 15.0102.060 2ω1s rad 0.309 ω

2

2

1IωK r

J 100rK

20.30921002

1rK

N; 18.0 ;m kg 2100 ;m 2.40 2 FIR0 s; 15.0 0 ωt

DF025 CHAPTER 8

117

Solution :

b. The angular displacement, for 15.0 s is given by

By applying the formulae of work done in rotational motion, thus

c. The average power supplied by the force is

τθW

20 2

1αttωθ

t

WPav

rad 2.32θ

2.3218.02.40W

N; 18.0 ;m kg 2100 ;m 2.40 2 FIR0 s; 15.0 0 ωt

22 15.0102.062

10 θ

RFθW

J 100W

15.0

100avP

W6.67avP

DF025 CHAPTER 8

118

At the end of this chapter, students should be able to: Define and use the formulae of angular momentum,

State and use the principle of conservation of angular momentum

Learning Outcome:

8.5 Conservation of angular momentum (1 hour)

IωL

DF025 CHAPTER 8

119

Conservation of angular momentumAngular momentum, is defined as the product of the angular velocity of a body

and its moment of inertia about the rotation axis.

OR

It is a vector quantity. Its dimension is M L2 T1

The S.I. unit of the angular momentum is kg m2 s1.

L

where

IL momentumangular : L

body a of inertia ofmoment : Ibody a oflocity angular ve : ω

is analogous to the mvp

DF025 CHAPTER 8

120

The relationship between angular momentum, L with linear momentum, p is given by vector notation :

magnitude form :

Newton’s second law of motion in term of linear momentum is

hence we can write the Newton’s second law in angular form as

and states that the vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum.

vmrprL

θmvrθrpL sinsin where

vrθ

with between angle the:axisrotation the toparticle thefrom distance : r

dt

pdFF nett

dt

Ldττ nett

DF025 CHAPTER 8

121

states that the total angular momentum of a system about an rotation axis is constant if no external torque acts on the system.

OR

Therefore

Principle of conservation of angular momentum

constantI

0 dt

Ldτ

0Ld

If the 0 τ

if L-LdL

fi LL

and

DF025 CHAPTER 8

122

A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s-1 about the rotation axis as shown in Figure 7.21. A 50 kg bag of sand falls onto the disc at the edge of the wooden disc.

Calculate,

a. the angular speed of the system after the bag of sand falling

onto the disc. (treat the bag of sand as a particle)

b. the initial and final rotational kinetic energy of the system.

Why the rotational kinetic energy is not the same?

(Use the moment of inertia of disc is )

Example 19 :

0ω

Before

R

After

R

Figure 7.21

2

2

1MR

DF025 CHAPTER 8

123

Solution :

a. The moment of inertia of the disc,

The moment of inertia of the bag of sand,

By applying the principle of conservation of angular momentum,

kg 50 kg; 200 ;s rad 4.0 ;m 3.00 10

bw mmωR

22 3.002002

1

2

1 RmI ww

1s rad 2.67 ω

ωIIωI bww 0

22 3.0050 RmI bb

fi LL

2m kg 900wI

2m kg 450bI

ω4509004.0900

DF025 CHAPTER 8

124

Solution :

b. The initial rotational kinetic energy,

The final rotational kinetic energy,

thus

It is because the energy is lost in the form of heat from the

friction between the surface of the disc with the bag of

sand.

kg 50 kg; 200 ;s rad 4.0 ;m 3.00 10

bw mmωR

220 4.0900

2

1

2

1 ωIK wir

22 2.674509002

1

2

1 ωIIK bwfr

J 7200irK

frir KK J 4812frK

DF025 CHAPTER 8

125

A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer.

Solution :

The answer is hard-boiled egg.

Example 20 :

DF025 CHAPTER 8

126

Solution :

Reason

Raw egg :

When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases because of

From the principle of conservation of angular momentum,

If the I is increases hence its angular velocity, will decreases.

Hard-boiled egg :

The position of the yolk of a hard-boiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.

2mrI

constantI

DF025 CHAPTER 8

127

Before After

0

A student sits on a freely rotating stool holding two weights, each of mass 3.00 kg as shown in Figure 7.22.

When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad s1. The moment of inertia of the student and stool is 3.00 kg m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. Determine

a. the new angular speed of the student.

b. the kinetic energy of the rotating

system before and after he pulls

the weights inward.

Example 21 :

Figure 7.22

DF025 CHAPTER 8

128

Before

0

After

ar ar

Solution : 210 m kg 3.00 ;s rad 0.750 ;kg 3.00

ssw Iωm

brbr

wmwm

DF025 CHAPTER 8

129

Solution :

a. The moment of inertia of the system initially is

The moment of inertia of the system finally is

By using the principle of conservation of angular momentum,

thus

wssi III 22bwbwssi rmrmII

22 bwssi rmII 22 m kg 9.001.003.0023.00 iI

22 awssf rmII 22 m kg 3.540.3003.0023.00 fI

;m kg 3.00 ;s rad 0.750 ;kg 3.00 210

ssw Iωmm 0.300 ;m 1.00 ab rr

1s rad 1.91 ω

ωIωI fi 0

fi LL

ω3.540.7509.00

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130

Solution :

b. The initial rotational kinetic energy is given by

and the final rotational kinetic energy is

202

1ωIK iir

20.7509.002

1irK

J 2.53irK

2

2

1ωIK ffr

21.913.542

1frK

J 6.46frK

;m kg 3.00 ;s rad 0.750 ;kg 3.00 210

ssw Iωmm 0.300 ;m 1.00 ab rr

DF025 CHAPTER 8

131

Exercise 8.5 :Use gravitational acceleration, g = 9.81 m s2

1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s1 relative to the Earth.

a. In the what direction and with what value of angular speed

does the turntable rotate?

b. How much work does the woman do to set herself and the

turntable into motion?

ANS. : 0.360 rad s1 ,U think; 99.9 J

DF025 CHAPTER 8

132

Exercise 8.5 :2. Determine the angular momentum of the Earth

a. about its rotation axis (assume the Earth is a uniform solid

sphere), and

b. about its orbit around the Sun (treat the Earth as a particle

orbiting the Sun).

Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m and is 1.5 x 108 km from the Sun.

ANS. : 7.1 x 1033 kg m2 s1; 2.7 x 1040 kg m2 s1

3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is )

ANS. : 4.71 x 106 kg m2 s1

2

12

1ML

DF025 CHAPTER 8

133

Linear Motion Relationship Rotational MotionSummary:

m

rv dt

d

dt

dva

dt

d

maF I

FsW W

dt

dsv

ra

sinrF

FvP P

mvp IL sinrpL

I2i

n

1iirmI

DF015 CHAPTER 8

134

THE END…Next Chapter…

CHAPTER 9 :Simple Harmonic Motion

Sports

Physics Chapter 8- Rotational of a Rigid Body