      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=mr 2 (units are kg m 2 )

 If these cylinders have the same mass, which will reach the bottom of the ramp 1 st ?

 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

 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.

cylindrical shell :

I



MR

2

solid cylinder :

I



1 2

MR

2

rod about center :

I



1 12

ML

2

rod about end :

I



1 3

ML

2

spherical shell :

I

solid sphere :

I

 

2 3

MR

2

2 5

MR

2 bicycle rim filled can of coke baton baseball bat basketball boulder

 For a given amount of torque applied to an object, its rotational inertia determines its rotational acceleration acceleration  the smaller the rotational inertia, the bigger the rotational

F net

 

ma



I



Same torque, different rotational inertia

Big rotational inertia

spins slow

Small rotational inertia

spins fast

Treat 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 kg  m 2 b) 6.7 rad/s 2 c) 4 m/s 2 d) 1.7 kg e) 12.5 m

A cylindrical space station of (R=12, M=3400 kg) has moment of inertia 0.75 MR 2 . Retro rockets are fired tangentially at the surface of space station and provide impulse of 2.9x10

4 N·s. 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) w = 0.948 rad/s b) a=10.8 rad/s 2

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.

I

 1 3

ml

2 90 o θ θ F y θ  + F x mg .

θ = 35 o   

I

    

o



3.1



mg

cos(  ) 2  

L

 1

mL

2 2 3 3

g

cos(  )  2

L

 

K K

  1 2

mv

2 , 1 2 w ) 2

v t K K rot

 1 2

mr

2 w 2 ,  1 2

I

w 2 

r

w

I



mr

2

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.

I

 1 3

ml

E

i

2 

E

f

U

g



K

R

.

θ = 35 o θ = 35 o

h



L

2   )

mgh



1 2

I

w 2

mg L

2



1 1

mL

2 w 2

2 3

at lowest point Answer: ω = 3.6 r/s  w 

L

  2

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 w

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.

E before U g

 

K T E after



K R mgh



v



R

w 1 2

mv

2  1 2

I

w 2

I sphere

@

cm

 2 3

mR

2

mgh



gh

 1 2

mv

2  1 2 ( 2 3

mR

2 )(

v

2

R

2 ) 1 2

v

2  1 3

v

2

v

 6 5

gh

 

7.7 m/s

E before U g

 

K T E after mgh

 1 2

mv

2

mgh

 1 2

mv

2

gh



v

 1 2

v

2 2

gh

 2(10)(5)  10 If you HAD NOT included the rotational kinetic energy, you see the answer is very much different.

 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

Rigid body

L



I

w

L



mvr



m

w

r

2 Point particle

Analogy between L and p

Angular Momentum Linear momentum L = I

w 

t =

D

L Conserved if no net outside torques p = mv Ft =

D

p Conserved if no net outside forces

Rank the following from largest to smallest angular momentum.

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

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.

a) What is the period of the new motion?

T F = T 0 /25 = 0.2 s b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer?

W = 7.11x10

5 J

The 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)

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.

Four 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 m 2 /s

The 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

Quantity

Position Displacement Velocity Acceleration 1 st kinematic 2 nd kinematic 3 rd kinematic Centripetal acceleration Inertia Kinetic Energy What causes acceleration Newton’s 2 nd Law Momentum

Linear

x (or y) Δx v=Δx/Δt a=Δv/Δt

v f

2

v f

  0

v t o

 1 2

at at

2 

v

0 2

a c



v

2

r

m KE trans = ½ mv 2 Force F net =ma p=mv

Rotational

Ɵ Δ Ɵ ω=Δ Ɵ /Δt=2π/T α=Δω/Δt w

f

 w  0 

t

 w

o t

 1 2 

t

2 w

f

2  w 0 2  2   

c

 w 2

r

I KE rot = ½ Iω 2 Torque τ net =Iα L=Iω or mωr 2

Connection

Δx=rΔ Ɵ v=rω a=rα τ=r*F