1. Unit 7 - Rotational Mechanics
Ch 8 – Kinematics +
Ch 9 – Dynamics

2. Part 3 – Rotational Dynamics
Book Chapter 9

3. Newton’s 2nd Law for Rotation
So far in this unit we learned about the rotational equivalent of force, which we called torque:
𝐹 → 𝜏=𝐹𝑟
And rotational kinematics and variables:
𝑥, 𝑣, 𝑎 → 𝜃, 𝜔,𝛼 when you divide by r
If we want the rotational equivalent of Newton’s 2nd Law:
𝐹=𝑚𝑎 → 𝜏=𝐼𝛼
But what is 𝐼?

4. 𝐼 – the “Moment of Inertia”
𝐼 is called the moment of inertia or rotational inertia
(Remember that mass is a measure of translational inertia!)
For a point mass (or an object in orbit):
𝐼=𝑚 𝑟 2 , where r is the radius of orbit.
For any object:
𝐼= 𝑚 𝑟 2 , where r is the distance to the axis of rotation.
Units for 𝐼: 𝑘𝑔∙ 𝑚 2

5. Calculating 𝐼
Sometimes 𝐼 is given, and sometimes we need to calculate it (or at least be able to compare to solve problems)
I depends on how the mass of an object is distributed:
For a hoop or ring, 𝐼=𝑚 𝑟 2
For a solid cylinder, 𝐼= 1 2 𝑚 𝑟 2 (note: length doesn’t matter!)
For a hollow sphere, 𝐼= 2 3 𝑚 𝑟 2
For a solid sphere, 𝐼= 2 5 𝑚 𝑟 2

6. Example:
Answer: i – greater about the left end. Masses that are farther away from the axis of rotation have more rotational inertia.

7. Acceleration
If we solve for α, we can compare how quickly something will rotate when a torque is applied.
Example: A 100 N force is applied perpendicular to the outside of a 2-kg sphere with radius 0.8 m. What will the angular acceleration be if…
A) the sphere is solid?
𝛼= 𝜏 𝐼 = 100∗ ∗2∗ ≈156 𝑟𝑎𝑑 𝑠 2
B) the sphere is hollow?
𝛼= 𝜏 𝐼 = 100∗ ∗2∗ ≈94 𝑟𝑎𝑑 𝑠 2 (Note: more inertia so harder to make move!)

8. Rotational Work and Energy
Since 𝑊=𝐹𝑑= 𝜏 𝑟 ∙𝜃𝑟, our rotational work can be calculated with the formula:
𝑊 𝑅 =𝜏𝜃 (unit: joule)
Similarly, our kinetic energy formula for translation motion (𝐾𝐸= 1 2 𝑚 𝑣 2 ) can be expressed for rotational motion:
𝐾𝐸 𝑅 = 1 2 𝐼 𝜔 2 (unit: joule)

9. Conservation of Energy
Since rotational energy is another type of mechanical energy, conservation of energy can be rewritten to include it:
𝐸 𝑜 = 𝐸 𝑓 → 𝑃𝐸 0 + 𝐾𝐸 0 + 𝐾𝐸 𝑅0 = 𝑃𝐸 𝑓 + 𝐾𝐸 𝑓 + 𝐾𝐸 𝑅𝑓
A classic problem involved different shapes rolling down an incline. In this case:
𝑃𝐸 𝑡𝑜𝑝 = 𝐾𝐸 𝑏𝑜𝑡𝑡𝑜𝑚 + 𝐾𝐸 𝑅𝑏𝑜𝑡𝑡𝑜𝑚
As the shapes roll down the incline, the potential energy is converted into kinetic energy, but it is shared between translational and rotational.
The objects with the least rotational inertia should have less rotational energy and hence MORE translational energy.

10. Answer: A=C>B>D For the same masses, the objects with more rotational inertia will have less translation energy (so speed). Since mass is in each type of energy, it won’t make a difference, so A=C.

11. Answer: B – below point A
Answer: B – below point A. Some of the potential energy from the top will have been converted into rotational energy as the sphere rolls. Less potential energy left means that the sphere will not go as high.

12. Angular Momentum
Our final rotational relationship with involve angular momentum (L).
𝑝=𝑚𝑣 → 𝐿=Iω (unit: 𝑘𝑔∙ 𝑚 2 𝑠 )
The Principle of Conservation of Angular Momentum says that the angular momentum of a system is conserved if the net average external torque acting on the system is zero.
𝐿 0 = 𝐿 𝑓

13. Angular Momentum
Why does a spinning ice skater spin faster if she bring her arms and legs in? Is angular momentum being conserved?
There is no external torque acting on the skater, so angular momentum should be conserved.
As the skater brings her arms and legs in, her moment of inertia decreases, so her speed increases

14. Answer: D. The change in momentum for the sphere is pf + pi
Answer: D. The change in momentum for the sphere is pf + pi. Angular momentum is mvr, ∆p*d is the change in angular momentum for the rod.