Rotational motionRotational motioni. Able to define and write equations for the angular

quantities in rotational motion.ii. Able to define and analyze rotational motion with

constant angular acceleration problems.iii. Able to state and define the formula for torque in

terms of the angular acceleration and the moment of inertia.

iv. Able to explain the effect of moment of inertia of an object undergoing rotational motion.

Table below shows the symbols used in linear and rotational motion.

Linear motion

QuantityRotational

motion

s θDisplacementDisplacement

u 0ωInitial velocityInitial velocity

v ωFinal velocityFinal velocity

a αAccelerationAcceleration

t tTimeTime

Comparison between linear and rotational motion with constant acceleration.

LINEAR MOTION ROTATIONAL MOTION

v u a t

21

2s u t a t

21 -

2s v t a t

2 2 2 v u a s

1

2s u v t

o t

21

2o t t

21 -

2t t

2 2 2 o s

1

2 o t

constanta constant

a

c ir c u lar p a th

p o in ts to w ar d s th ec en tr e o f th e c ir c le

Relation between linear & rotational motion

a) The position : where in radians

b) The speed : where : tangential / linear velocity ( in ms-1 ) : angular velocity ( in rads-1 )

c) The acceleration : where : tangential / linear acceleration ( in ms-2 ) : angular acceleration ( in rads-2 )

s r

tv r

ta r

tv

ta

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 :Solution :

a. The initial angular velocity is

and the angular acceleration of the wheels is given by

Example 1

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

Solution :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

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 :Solution :

Example 2

,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

Solution :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

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 :Solution :

a. Given t =4.00 s, thus

Example 3

,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

Solution :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

Solution :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

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 :Solution :

The radius of the coin is

Example 4

m 102.40 2d

1s rad 18 0ω

s

2s rad 1.90 α

1s rad 0 ω

m 101.202

2d

r

Solution :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

A wheel turning with angular speed of 30 rps is brought to rest with constant acceleration. It turns 60 revolutions before it stops.

a)What is its angular acceleration?b)What time elapsed before it stops?

TORQUE

The tendency of a force to rotate an object about an axis is measured by torque,

where is moment inertia

is angular acceleration

I

I

Figure below 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 the sum of the products of the mass of each particle and the square of its respective distance from the rotation axisand the square of its respective distance from the rotation axis.

Moment of inertia, I

1m

2mnm

3m

1r2r

3rnr

O

OR

It is a scalar quantityscalar quantity. Moment of inertia, Moment of inertia, II in the rotational kinematics is analogousanalogous to the

mass, mass, mm in linear kinematics. The dimensiondimension of the moment of inertia is M LM L22. The S.I. unitS.I. unit of moment of inertia is kg mkg m22. The factorsfactors which affect the moment of inertia, I of a rigid body:

a. the massmass of the body,b. the shapeshape of the body,c. the positionposition of the rotation axisrotation 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

22

Here, v 0. The direction of v is changing.

If v 0, then a 0. The net force cannot be zero.

Consider an object moving in a circular path of radius r at constant speed.

x

y

v

v

v

v

5. Centripetal /radial acceleration and Centripetal force

The velocity is tangent to its path.

23

Conclusion:

to move in a circular path, an object must have a nonzero net force acting on it.

From Newton’s Second Law, F = ma, where Fnet acting on the system in uniform circular motion is given by:

Fc = mac

Centripetal/ radial accelerationCentripetal force

For an object moving in uniform circular motion, the acceleration is radially inward,and in the same direction as the centripetal force.

24

The magnitude of the radial acceleration is:

vrr

var 2

2

For motion with uniform circular motion, the kinematic equations:

uts t0

Linear Angular

where / 0 2srad

An object travels with uniform circular motion at an orbital speed of 3.0 m/s and radius of 1.5 m. The object then experiences a force for 2.0 s resulting in an angular acceleration of 0.40 rad/s2. The object remains in the same circular orbit. Calculate

a)the acceleration of the object before it experiences the force

b) the angular speed of the object after the 2 s interval

c) the angular displacement undergone by the object after the 2 s interval

d) the acceleration of the object after the 2 s interval