1. Rotational Inertia & Kinetic Energy

2. Linear & Angular   Linear Angular Displacement x θ Velocity v
Acceleration a 
Inertia m I KE
½ mv2
½ I2 N2
F = ma
 = I
Momentum
P = mv
L = I

3. Rolling Motion
If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

4. Rolling Motion
We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

5. Rolling Motion
We may also consider rolling motion at any given instant to be a pure rotation at rate w about the point of contact of the rolling object.

6. A Rolling Tire
A car with tires of radius 32 cm drives on a highway at a speed of 55 mph.
(a) What is the angular speed w of the tires?
(b) What is the linear speed vtop of the top to the tires?

7. Rotational Kinetic Energy
Consider a mass M on the end of a string being spun around in a circle with radius r and angular frequency w
Mass has speed v = w r
Mass has kinetic energy
K = ½ M v2
K = ½ M w2 r2
Rotational Kinetic Energy is energy due to circular motion of object. M 24

8. Rotational Inertia I Ktran = ½ m v2 Linear Motion
Tells how much “work” is required to get object spinning. Just like mass tells you how much “work” is required to get object moving.
Ktran = ½ m v2 Linear Motion
Krot = ½ I w Rotational Motion 13

9. Inertia Rods Two batons have equal mass and length.
Which will be “easier” to spin?
A) Mass on ends
B) Same
C) Mass in center
I = S m r2 Further mass is from axis of rotation, greater moment of inertia (harder to spin)

10. Inertia of a Dumbbell
Use the definition of moment of inertia to calculate that of a dumbbell-shaped object with two point masses m separated by a distance of 2r and rotating about a perpendicular axis through their center of symmetry.
If the rod has mass then +Irod=1/12

11. Nose to the Grindstone
A grindstone of radius r = m is being used to sharpen an axe.
If the linear speed of the stone is 1.50 m/s and the stone’s kinetic energy is 13.0 J, what is its moment of inertia I ?

12. Moment of Inertia of a Hoop
All of the mass of a hoop is at the same distance R from the center of rotation, so its moment of inertia is the same as that of a point mass rotated at the same distance.

13. Moments of Inertia

14. I is Axis Dependent

15. Rolling Objects

16. Like a Rolling Disk
A 1.20 kg disk with a radius 0f 10.0 cm rolls without slipping. The linear speed of the disk is v = 1.41 m/s.
(a) Find the translational kinetic energy.
(b) Find the rotational kinetic energy.
(c) Find the total kinetic energy.

17. Preflight: Rolling Race (Hoop vs Cylinder)
A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest KE at bottom?
A) Hoop B) Same C) Cylinder
20% % %

18. Preflight: Rolling Race (Hoop vs Cylinder)
A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest speed at the bottom of the ramp?
A) Hoop B) Same C) Cylinder
I = MR2
I = ½ MR2

19. Rolling Down an Incline

20. Spinning Wheel mgh=1/2mv2(1+I/mR2)
A block is attached to a string around a pulley. The string is pulled rising the block at a velocity v spinning the pulley with a rotational velocity w .
To what height h does the block rise?
mgh=1/2mv2(1+I/mR2)

21. A Bowling Ball
A bowling ball that has an 11 cm radius and a 5.0 kg mass is rolling without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill.
Model the bowling ball as a uniform sphere and calculate h.

22. Torque and Energy Remember Torque=Fd=Iα A torque makes an acceleration
So with an initial torque you can spin something up and that can roll to do work.

23. Example
A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds.
What is the initial ω?
If the coefficient of rolling friction is μ=.05 use work to find how far the wheel rolls.

24. Example
A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds.
What is the initial ω?
ω=αt τ=Iα so α= τ/I
ω=t (τ/I) Idisk= 1/2mr2
ω=t (τ/(1/2mr2))
ω=5 (5/(1/2*1*.52)) = 200 rad/s2

25. Example
A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds.
If the coefficient of rolling friction is μ=.05 use work to find how far the wheel rolls.
ω=200 rad/s KE=1/2mv2 + 1/2Iω2
v= ωr Idisk= 1/2mr2 KE=1/2m(ωr)2 + ½(1/2mr2)ω2
KE=3/4m(ωr)2 =.75*1*(200*.5)2 = 7,500 J