41
Rotational Motion & Angular Momentum

Rotational Motion & Angular Momentum · Angular Momentum r v m Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol

  • Upload
    others

  • View
    49

  • Download
    5

Embed Size (px)

Citation preview

Rotational Motion &

Angular Momentum

Rotational Motion

Every quantity that we have studied with

translational motion has a rotational

counterpart

TRANSLATIONAL ROTATIONAL

Displacement x Angular Displacement

Velocity v Angular velocity

acceleration a Angular acceleration

Mass m Inertia I

Momentum p Angular Momentum L

Force F Torque

Angular Position

r

s

Arclength

Radius (from axis of

rotation)

Measured in radians – all angular quantities will be measured in

radians (NOT degrees)

• In translational motion,

position is represented by a

point, such as x.

• In rotational motion,

position is represented

by an angle, such as ,

and a radius, r.

x

linear

0 5

x = 3

r

0

p/2

p

3p/2angular

Angular Position

Displacement

• Linear displacement

is represented by the

vector Dx.

• Angular displacement is represented by D, which is not a vector, but behaves like one for small values.

x

linear

0 5

Dx = 4

D 60

0

p/2

p

3p/2angular

Angular Displacement

0DWhich direction is positive (by

convention)?

Positive – Counterclockwise

Negative - Clockwise

Compare to

D𝒙 = 𝒙 − 𝒙𝒐

EXAMPLE: (a) What is the angular position, , if we go

around a circle two times?

Ans. = s

r=

2(2p𝑟)r

= 4π

(b) Say you go a quarter turn more, what is D?

Ans. D = o = 9𝜋

2− 4π =

𝜋

2(c) What is the arclength covered in total between

Probls. (a) and (b) if the radius of the circle is 3 m?

Ans. s = r = (3 m)(4p + 𝜋

2) =3m(

9𝜋

2) =

27𝜋

2m

P.O.D. 1: A Boy

on a horse is

hunting a goose

around a strange

world of radius 25

m. The angular

separation

between the

hunter/hunted is a

constant 5. What

is the angular

distance (in m)

between the

boy/horse and the

goose?

Speed and velocity

• The same particle rotates with an avg. angular velocity given by

= D𝛉

D𝐭• Linear (Tangential) and angular

speeds are related by the equation

v = r

rD

s

vT

vT

Angular Velocity

tavg

D

D

Units – rad/s, rev/s

Direction – Same as

displacement

(positive is

counterclockwise,

negative is

clockwise)

Compare to

𝒗𝒂𝒗𝒈 =∆𝒙

∆𝒕

EXAMPLE: A space station ring of radius 500 m

spins twice in 12 minutes.

(a) Find its angular velocity

(in rads/s).

(b) Find the linear

(tangential) velocity (in

m/s) of a point on the outer

edge of the ring.

Ans. (a) Twice means two revolutions = 2(2p) = 4 p radians

= 𝜃

𝑡=

4𝜋

12 min × 60 Τ𝑠𝑒𝑐𝑚𝑖𝑛

= 0.017 rad/sec

(b) tangential velocity, vT = r = 0.017 (500 m) = 8.73 m/s

P.O.D. 2: A figure skater spins through five

revolutions in a time of 2.4 s.

(a) Find her angular velocity (in rads/s).

(b) Find the linear (tangential) velocity (in m/s) of her

foot if it is 0.3 meters away from her body.

Acceleration

• The same particle rotates with an avg. angular acceleration given by

= DD𝐭

• Linear (Tangential) and angular accelerations are related by the equation

a = r

rD

s

vT

vT

Angular Acceleration

• Angular

Acceleration is

how the angular

velocity changes

with time

tavg

D

D

Units – rad/s2

Compare to

𝒂𝒂𝒗𝒈 =∆𝒗

∆𝒕

Centripetal Acceleration

rr

vac

22

A pendulum is swinging back and

forth.

At the bottom of the swing the

force of gravity is pulling it

downwards but it doesn’t fall down.

This means there must be a force

pulling upwards to balance it out.

This is the centripetal force.

Since F = ma, the center-seeking

acceleration is called centripetal

acceleration and is given by:

Fc

Fg

Sample Problem

A compact disk rotates about an

axis according to the formula:

(t) = t2 – 6

(a) What is the linear speed of a point 20 cm from the

center at t = 5 s?

(b) What is the linear acceleration at 0.5 m at t = 5 s?Ans.

(a) To find the linear speed, use the relationship, where (5)

we obtained from the function (t) = t2 – 6 (5)= 52 –6=19v = r = (19 rad/s)(0.20 m) = 3.8 m/s.

(b) To find the linear acceleration, use the relationship

ac = 2r = (19 rad/s)2(0.20 m) = 72.2 m/s2

P.O.D. 3: Diana Prince starts rotating with an angular

velocity of 5 rad/s, where the negative sign indicates a

clockwise rotation, to transform into Wonder Woman.

After 5 seconds she has completed the transformation

and her angular velocity is 2 rad/s.

(a) Find Wonder

Woman’s angular

acceleration (in

rad/s2).

(b) Find the centripetal

acceleration (in m/s2)

of her indestructible

bracelets (at = 2

rad/s) if they are 0.8

m away from her

body when she is

spinning.

Constant Angular Acceleration

• Our kinematics equations have angular

equivalents

• Just as with their linear counterparts, these only

work for constant acceleration

First Kinematic Equation

• x = vot + ½ at2 (linear form)

– Substitute angle for position.

– Substitute angular velocity for linear velocity.

– Substitute angular acceleration for linear

acceleration.

= ot + ½ t2 (angular form)

Second Kinematic Equation

• v = vo + at (linear form)

– Substitute angular velocity for linear velocity.

– Substitute angular acceleration for linear

acceleration.

= o + t (angular form)

Sample Problem

A bicycle starts from rest and for 10.0

s has a constant linear acceleration of

0.8 m/s2 to the right. During this

period, the tires do not slip. The

radius of the tires is 0.50 m. At the

end of the 10.0 s interval what is the

angle through which each wheel has

rotated?

Ans. The angular acceleration can be found from the formula

a = r 𝜶 =𝒂

𝒓 𝜶 =

𝟎.𝟖𝒎

𝒔𝟐

𝟎.𝟓𝟎 𝒎= 1.6 rad/s2

The angular acceleration should be negative because the tire

spins clockwise.

To find the angular displacement:

(t) = ot + ½t2 = 0(10 s) + ½(1.6 rad/s2)(10 s)2 = 80 rad

P. O. D. 4: an extreme diver rotates at an angular velocity of +3.75 rad/s

while doing his first set of flips. When he does his second set of flips, he

accelerates and reaches a greater angular velocity of +4.4 rads/s. If his

angular acceleration is 1.7 rad/s2...

(a) How long does this flipping part of the dive take (in s)?

(b) Find his angular displacement after the time found in (a).

ANGULAR

MOMENTUM!

Angular Momentum

r

v

m

Angular momentum depends on linear momentum and the distance

from a particular point. It is a vector quantity with symbol L. If rand v are then the magnitude of angular momentum w/ resp. to

point Q is given by L = rp = mvr. In this case L points out of the

page. If the mass were moving in the opposite direction, L would

point into the page.

The SI unit for angular momentum

is the kg m2 / s. (It has no special

name.) Angular momentum is a

conserved quantity. A torque is

needed to change L, just a force is

needed to change p. Anything

spinning has angular has angular

momentum. The more it has, the

harder it is to stop it from spinning.

Q

Angular Momentum: General

Definition

If r and v are not then the angle between these two vectors must

be taken into account. The general definition of angular momentum is

given by a vector cross product:

L = r pThis formula works regardless of the angle. From cross products, the magnitude of the

angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the

right-hand rule, L points out of the page. If the mass were moving in the opposite

direction, L would point into the page.

r

v

m

Q

Moment of Inertia vs. Angular Momentum

Any moving body has inertia. (It wants to keep moving at constant

v.) The more inertia a body has, the harder it is to change its linear

motion. Rotating bodies possess a rotational inertia called the

moment of inertia, I. The more rotational inertia a body has, the

harder it is change its rotation. For a single point-like mass w/ respect

to a given point Q, I = mr 2.

I = mr 2

m

r

For a system, I = the sum of each mass

times its respective distance from the

point of interest.

r2

m2

r1

m1

I = mi ri2 = m1 r1

2 + m2r22

Q

Q

Moment of Inertia of various shapes

Moment of Inertia ExampleTwo merry-go-rounds have the same mass and are spinning with the

same angular velocity. One is solid wood (a disc), and the other is a

metal ring. Which has a bigger moment of inertia relative to its center

of mass?

mm

r

r

Ans. I is independent of the angular speed. Since their masses and

radii are the same, the ring has a greater moment of inertia. This is

because more of its mass is farther from the axis of rotation. Since I

is bigger for the ring, it would more difficult to increase or decrease its

angular speed.

Torque & Angular Acceleration

Newton’s 2nd Law, as you know, is Fnet = ma

The 2nd Law has a rotational analog: net = I

A force is required for a body to undergo acceleration. A “turning force”

(a torque) is required for a body to undergo angular acceleration.

The bigger a body’s mass, the more force is

required to accelerate it. Similarly, the

bigger a body’s rotational inertia, the more

torque is required to accelerate it angularly.

Both m and I are measures of a body’s

inertia

(resistance to change in motion).

Example: The torque of an Electric Saw Motor

The motor in an electric saw brings the circular

blade up to the rated angular speed of 80.0 rev/s

in 240.0 rev. One type of blade has a moment of

inertia of 1.41 x 10-3 kgm2. What net torque

(assumed constant) must the motor apply to the

blade?

Ans. First we need to convert our values into rad/s for calculation

purposes.

o t240 rev x 2p = 1508 rads ? 80 revs/s x 2p = 503 rad/s 0 rad/s ?

We can find the angular acceleration from 2 = o2 + 2

Solving for : = 2−o

2

2= (503 rad/s)2−02

2(1508 rad/s)= 83.89 rad/s2

= I = 1.41 x 10-3 kgm2 83.89 rad/s2 = 0.118 Nm

PROBLEM 6: A Chinese star of mass 0.025 kg and radius 0.03 m is

thrown by Bruce Lee at his adversary. The Chinese star is thrown

from rest. If its final angular velocity is 15 revs/s after 3 sec,

(a) find the angular acceleration of the Chinese star.

(b) Find the torque of the Chinese star (Assume the Chinese star is

hoop-shaped).

Linear Momentum vs. Angular MomentumIf a net force acts on an object, it must accelerate, which means its

momentum must change. Similarly, if a net torque acts on a body, it

undergoes angular acceleration, which means its angular momentum

changes. Recall, angular momentum’s magnitude is given by

L = mvr

r

v

m

So, if a net torque is applied, angular velocity must

change, which changes angular momentum.

Proof: net = r Fnet = r m a

= r mDvt

= DLt

So net torque is the rate of change of angular momentum, just as net

force is the rate of change of linear momentum.continued on next slide

(if v and r are perpendicular)

Linear & Angular Momentum (cont.)

Here is yet another pair of similar equations, one linear,

one rotational. From the formula v = r , we get

L = mv r = m ( r) r = m r2 = I

This is very much like p = mv, and this is one reason I is

defined the way it is.

In terms of magnitudes, linear momentum

is inertia times speed, and angular

momentum is rotational inertia times

angular speed.

L = I

p = m v

Comparison: Linear & Angular Momentum

Linear Momentum, p

• Tendency for a mass to continue

moving in a straight line.

• Parallel to v.

• A conserved, vector quantity.

• Magnitude is inertia (mass)

times speed.

• Net force required to change it.

• The greater the mass, the greater

the force needed to change

momentum.

Angular Momentum, L

• Tendency for a mass to continue

rotating.

• Perpendicular to both v and r.

• A conserved, vector quantity.

• Magnitude is rotational inertia

times angular speed.

• Net torque required to change it.

• The greater the moment of

inertia, the greater the torque

needed to change angular

momentum.

Example: Spinning Ice SkaterSuppose Mr. Stickman is sitting on a stool that swivels holding a pair of

dumbbells. His axis of rotation is vertical. With the weights far from that axis,

his moment of inertia is 600 kgm2 and he is spinning at an angular velocity of

20 rad/s. When he pulls his arms in as he’s spinning, the weights are closer to the

axis, so his moment of inertia gets to

400 kgm2. What will be his angular

velocity at this moment?

Ans.

I = L = I 600 kgm2(20 rad/s) = 400 kgm2

12,000 = 400

30 rad/s =

PROBLEM 7: An artificial

satellite (m = 1500 kg) is placed

into an elliptical orbit about the

earth. Telemetry data indicate

that its point of closes approach

(called the perigee) is rp = 8.37

x 106 m from the center of the

earth, while its point of greatest

distance (called the apogee) is

rA = 25.1 x 106 m from the

center of the earth. The speed

of the satellite at the perigee is

vp = 8450 m/s.

(a) Find its speed vA at the

apogee.

(b)Find its angular momentum

at any point in its orbit.

Rotational Kinetic Energy

• A particle in a rotating object has rotational kinetic energy:

Ki = ½ mivi2, where vi = i r (tangential velocity)

2 2

2 2 2

1

2

1 1

2 2

R i i i

i i

R i i

i

K K m r

K m r I

The whole rotating object has a rotational kinetic energy given by:

Rotational Kinetic

Energy example:

A thin walled hollow cylinder (mass

= mh, radius = rh) and a solid

cylinder ( mass = ms, radius = rs)

start from rest at the top of an

incline. Both cylinders start at the

same vertical height ho. All heights

are measured relative to an

arbitrarily chosen zero level thatpasses through the center of mass of a cylinder when it is at the

bottom of the incline. Ignoring energy losses due to retarding

forces, determine which cylinder has the greatest translational

speed upon reaching the bottom.

Rotational Kinetic Energy example (cont.):

Ans. At the top of the incline the cylinder have only gravitational potential

energy. At the bottom of the incline this energy has converted into

translational kinetic and rotational kinetic energy.

Ein = Eout

GPEin = TKEout + RKEout

mgh = ½mvf2 + ½If

2

The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟

𝐫

Substituting the given values and for the angular velocity:

mhgho = ½mhvf2 + ½I(

𝐯𝐟

𝐫𝒉)2

For the hollow cylinder, the moment of inertia is given by: I = mr2

Substituting: mhgho = ½mhvf2 + ½(mhrh

2)(vf

rℎ)2

Simplifying: gho = ½vf2 + ½rh

2(𝐯𝐟

𝐫𝒉)2

gho = ½vf2 + ½rh

2𝐯𝒇𝟐

𝒓𝒉𝟐

gho = ½vf2 + ½𝐯𝒇

𝟐

gho = vf2

𝐠𝐡𝐨= vf

Rotational Kinetic Energy example (cont.):

Ans. At the top of the incline the cylinder have only gravitational potential

energy. At the bottom of the incline this energy has converted into

translational kinetic and rotational kinetic energy.

Ein = Eout

GPEin = TKEout + RKEout

mgh = ½mvf2 + ½If

2

The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟

𝐫

Substituting the given values and for the angular velocity:

msgho = ½msvf2 + ½I(

𝐯𝐟

𝐫𝒔)2

For the solid cylinder, the moment of inertia is given by: I = ½mr2

Substituting: msgho = ½msvf2 + ½(½msrs

2)(vf

r𝑠)2

Simplifying: gho = ½vf2 + ½ ½rs

2(𝐯𝐟

𝐫𝒔)2

gho = ½vf2 + ¼rs

2𝐯𝒇𝟐

𝒓𝒔𝟐

gho = ½vf2 + ¼𝐯𝒇

𝟐

gho = ¾vf2

𝟒

𝟑𝐠𝐡𝐨= vf

PROBLEM 8: