ROTATIONAL DYNAMICS

ROTATIONAL INERTIA (MOMENT OF INERTIA) Remember that inertia is the resistance an

object has to movement, it depends solely on mass

Rotational inertia measures the amount of torque it takes to get an object rotating, in other words it is the resistance of an object to accelerate angularly

It depends not only on the mass of the object, but where the mass is relative to the hinge or axis of rotation

The rotational inertia is bigger if more mass is located farther from the axis.

Moment of Inertia variable is I In general I=mr2 (units are kg m2)

EXAMPLE If these cylinders have the same mass,

which will reach the bottom of the ramp 1st?

The solid one! It has less moment of inertia because its mass is evenly distributed and the hollow one has it mass distributed farther away from its rotational axis

ANOTHER EXAMPLE Which of these two rods will be harder to

pick up? To spin?

They will be the same to pick up because their mass is the same, but the one on the left will be harder to spin because its mass is located farther from axis of rotation.

MOMENTS OF INERTIA – IF NEEDED FORMULAS WILL BE GIVEN

bicycle rim

filled can of coke

baton

baseball bat

basketball

boulder

cylindrical shell : I MR2

solid cylinder : I 1

2MR2

rod about center : I 1

12ML2

rod about end : I 1

3ML2

spherical shell : I 2

3MR2

solidsphere : I 2

5MR2

HOW FAST DOES IT SPIN? For a given amount of torque applied to

an object, its rotational inertia determines its rotational acceleration the smaller the rotational inertia, the bigger the rotational acceleration

netF ma

I

Big rotationalinertia

Small rotationalinertia

Same torque,different

rotational inertia

spinsslow

spinsfast

EXAMPLETreat the spindle as a solid cylinder.

a) What is the moment of Inertia of the spindle?

b) If the tension in the string is 10N, what is the angular acceleration of the wheel?

c) What is the acceleration of the bucket?

d) What is the mass of the bucket?

e) How far has the bucket dropped after 2.5 sec?

a) 0.9 kgm2 b) 6.7 rad/s2 c) 4 m/s2 d) 1.7 kg e) 12.5 m

EXAMPLE

a) What is the angular velocity of the space station after the rockets have finished firing?b) What is the centripetal acceleration at the edge of the space station?

A cylindrical space station of (R=12, M=3400 kg) has moment of inertia 0.75 MR2. Retro-rockets are fired tangentially at the surface of space station and provide impulse of 2.9x104 N·s.

a) w= 0.948 rad/s b) a=10.8 rad/s2

A 4 m beam with a 30 kg mass is free to rotate on a hinge. It is attached to a wall with a horizontal cable. The cable is then cut, find the initial angular acceleration of the beam.

mgcos()L

2

1

3mL2

. θ = 35o

mg Fx

Fy

+

90o-θ θ

θ

23 10 cos(35 )3.1 /

2 4

o

rad s

I

I 1

3ml2

3gcos()

2L

ROTATIONAL KINETIC ENERGY2

2

2 2 2

2

1 ,21 ( )21 ,2

12

t

rot

K mv v r

K m r

K mr I mr

K I

A 4 m beam with a 30 kg mass is free to rotate on a hinge. It is attached to a wall with a horizontal cable. The cable is then cut, find the angular velocity when the beam is horizontal.

E i E f

Ug KR

mgh12I 2

. θ = 35o

I 1

3ml2

2 2

2

1 1sin( )

2 2 3

3 sin( )1.9 /

Lmg mL

gr s

L

h L

2sin()

Find ω when beam is at lowest point

Answer: ω = 3.6 r/s

θ = 35o

Rolling Motion

Many rotational motion situations involve rolling objects.

Rolling without slipping involves both rotation and translation so you need to account for both rotational and translational kinetic energy.

Friction between the rolling object and the surface it rolls on is static, because the rolling object’s contact point with the surface is always instantaneously at rest.

this point on the wheel is instantaneously at rest if the wheel does not slip (slide)

The point of contact of the object with the surface is the axis of rotation

DEMO OF PULLING SPOOL

EXAMPLE

2 2

2@

22 2

2

2 2

1 12 2

23

1 1 2( )( )2 2 3

1 12 3

6 6 (10)(5)5 5

before after

g T R

sphere cm

E E

U K K

mgh mv I

v R I mR

vmgh mv mR

R

gh v v

v gh

2

2

2

12

12

12

2 2(10)(5) 10 /

before after

g T

E E

U K

mgh mv

mgh mv

gh v

v gh m s

In the past, everything was SLIDING. Now the object is rolling and thus has MORE energy than normal. So let’s assume the ball is like a thin spherical shell and was released from a position 5 m above the ground. Calculate the velocity at the bottom of the incline.

7.7 m/s

If you HAD NOT included the rotational kinetic energy, you see the answer is very much different.

EXAMPLE A bowling ball with a radius of .3 m and

mass of 4.6 kg is held by a wire. The wire is swung back so that it is now .5 m off the floor. The string is cut as the bowling ball makes contact with the floor. The resulting distance it travels in 5 seconds is 6 m. What is its rotational kinetic energy and speed?

19.7 J and 7.2 rad/s

L I

L mvr mr2

ANGULAR MOMENTUM (L)

Analogy between L and p

Angular Momentum Linear momentum

L = Iw p = mv

t t = DL Ft = Dp

Conserved if no netoutside torques

Conserved if no net outside forces

Rigid body

Point particle

EXAMPLERank the following from largest to smallest angular momentum.

EXAMPLE

A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

= 2.71 rad/s

EXAMPLE

a) What is the period of the new motion?

b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer?

Two twin ice skaters separated by 10 meters skate without friction in a circle by holding onto opposite ends of a rope. They move around a circle once every five seconds. By reeling in the rope, they approach each other until they are separated by 2 meters.

TF = T0/25 = 0.2 s

W = 7.11x105 J

EXAMPLEThe figure below shows two masses held together by a thread on a rod that is rotating about its center with angular velocity, ω. If the thread breaks, what happens to the system's (a) angular momentum and (b) angular speed. (Increase, decrease or remains the same)

EXAMPLE

A 50 kg figure skater rotates with her arms out at 2 rev/s, this gives her a radius of .7m. She then pulls her arms in which gives her a radius of 0.25 m. Find the final speed in rev/s.

EXAMPLEFour identical masses rotate about a common axis, 1.2 meters from the center. Each mass is 2.5 kg, and the system rotates at 2 rad/sec. The rods connecting them are assumed to be massless. Find the total angular momentum of this system.

1.2 m

2.5 kg

90.5 kg m2/s

EXAMPLE CONTINUEDThe four masses are then pulled toward the center until their radii are 0.5 meters. This is done in such a manner that no external torque acts on the system. What is the new angular speed of the system?

36.2 rad/s

REMEMBER OUR TABLE???Quantity Linear Rotational Connection

Position

Displacement

Velocity

Acceleration

1st kinematic

2nd kinematic

3rd kinematic

Centripetal acceleration

Inertia

Kinetic Energy

What causes acceleration

Newton’s 2nd Law

Momentum

x (or y) Ɵ

Δx ΔƟ Δx=rΔƟ

v=Δx/Δt ω=ΔƟ/Δt=2π/T v=rω

a=Δv/Δt α=Δω/Δt a=rα

0fv v at 0f t

21

2ox v t at 21

2ot t

2 20 2fv v a x 2 2

0 2f 2

c

va

r 2

c r

m I

KEtrans= ½ mv2 KErot= ½ Iω2

Force Torque τ=r*F

Fnet=ma τnet=Iα

p=mv L=Iω or mωr2