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1Prof. Sergio B. MendesSummer 2018
Chapter 10 of Essential University Physics, Richard Wolfson, 3rd Edition
Rotational Motion
2Prof. Sergio B. MendesSummer 2018
We’ll look for a way to describe the combined (rotational) motion
3Prof. Sergio B. MendesSummer 2018
Angle Measurements
When you’re not sure, always use radians to describe angles !!
2 𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 360° = 1 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝜃𝜃 ≡𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
4Prof. Sergio B. MendesSummer 2018
�𝜔𝜔 ≡∆𝜃𝜃∆𝑟𝑟
𝜔𝜔 ≡ lim∆𝑡𝑡→0
∆𝜃𝜃∆𝑟𝑟
=𝑟𝑟𝜃𝜃𝑟𝑟𝑟𝑟
Angular Velocity
𝜔𝜔 =𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟
Average:
Instantaneous:
Units:
5Prof. Sergio B. MendesSummer 2018
𝜔𝜔 ≡𝑟𝑟𝜃𝜃𝑟𝑟𝑟𝑟 =
1𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟 = 𝜔𝜔 𝑟𝑟
=𝑟𝑟𝑟𝑟
Angular Velocity and Linear Velocity
𝜃𝜃 ≡𝑟𝑟𝑟𝑟
6Prof. Sergio B. MendesSummer 2018
Example 10.1
𝑟𝑟 = 𝜔𝜔 𝑟𝑟𝑟𝑟 = 28 𝑚𝑚
𝜔𝜔 = 21 𝑟𝑟𝑟𝑟𝑚𝑚
𝑟𝑟 = 62 𝑚𝑚/𝑟𝑟
𝜔𝜔 = 2.2 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟
7Prof. Sergio B. MendesSummer 2018
�𝛼𝛼 ≡∆𝜔𝜔∆𝑟𝑟
𝛼𝛼 ≡ lim∆𝑡𝑡→0
∆𝜔𝜔∆𝑟𝑟
=𝑟𝑟𝜔𝜔𝑟𝑟𝑟𝑟
Angular Acceleration
𝛼𝛼 =𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟2
𝛼𝛼 =𝑟𝑟𝜔𝜔𝑟𝑟𝑟𝑟 =
𝑟𝑟 ⁄𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟
𝜔𝜔 =𝑟𝑟𝑟𝑟
=1𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
=1𝑟𝑟𝑟𝑟𝑡𝑡
𝑟𝑟𝑡𝑡 = 𝛼𝛼 𝑟𝑟 𝑟𝑟𝑟𝑟 =𝑟𝑟2
𝑟𝑟= 𝑟𝑟 𝜔𝜔2
Average:
Instantaneous:
Units:
8Prof. Sergio B. MendesSummer 2018
Linear Motion Analog to Rotation
9Prof. Sergio B. MendesSummer 2018
Example 10.2
𝛼𝛼 = −0.12 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟2
𝜔𝜔𝑜𝑜 = 2.2 𝑟𝑟𝑟𝑟𝑟𝑟/𝑟𝑟
𝜔𝜔2 = 𝜔𝜔𝑜𝑜2 + 2 𝛼𝛼 𝜃𝜃 − 𝜃𝜃𝑜𝑜
𝜔𝜔 = 0
𝜃𝜃 − 𝜃𝜃𝑜𝑜 = 20 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
= 3.2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
10Prof. Sergio B. MendesSummer 2018
Torque (the analog of force for rotational motion)
𝜏𝜏 ≡ 𝐹𝐹 𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 𝜃𝜃
𝐹𝐹⊥ = 𝐹𝐹 𝑟𝑟𝑟𝑟𝑟𝑟 𝜃𝜃
11Prof. Sergio B. MendesSummer 2018
Example 10.3
𝜏𝜏 = 𝐹𝐹 𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 𝜃𝜃
𝐹𝐹 = ? ?
𝑟𝑟 = 45 𝑠𝑠𝑚𝑚
𝜃𝜃 = 67°
𝜏𝜏 = 95 𝑁𝑁 𝑚𝑚
𝐹𝐹 = 230 𝑁𝑁
12Prof. Sergio B. MendesSummer 2018
Newton’s Second Lawfor Rotational Motion
𝑟𝑟𝑡𝑡 = 𝛼𝛼 𝑟𝑟
𝐹𝐹⊥ = 𝑚𝑚 𝑟𝑟𝑡𝑡
𝜏𝜏 = 𝐹𝐹⊥ 𝑟𝑟
𝜏𝜏 = 𝐼𝐼 𝛼𝛼
𝐼𝐼 ≡ 𝑚𝑚 𝑟𝑟2 Moment of Inertia or Rotational Inertia
= 𝑚𝑚 𝛼𝛼 𝑟𝑟
= 𝑚𝑚 𝛼𝛼 𝑟𝑟2
13Prof. Sergio B. MendesSummer 2018
𝐼𝐼 ≡�𝑖𝑖
𝑚𝑚𝑖𝑖 𝑟𝑟𝑖𝑖2
Moment of Inertia along a rotation axis
14Prof. Sergio B. MendesSummer 2018
Example 10.4
𝐼𝐼 =58𝑚𝑚 𝐿𝐿2
𝑚𝑚 = 0.64 𝑘𝑘𝑘𝑘
𝐿𝐿 = 85 𝑠𝑠𝑚𝑚
= 0.29 𝑘𝑘𝑘𝑘 𝑚𝑚2
15Prof. Sergio B. MendesSummer 2018
Moment of Inertia for a continuous distribution of mass
𝐼𝐼 ≡�𝑖𝑖
𝑚𝑚𝑖𝑖 𝑟𝑟𝑖𝑖2 𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚
16Prof. Sergio B. MendesSummer 2018
Example 10.5
𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚
𝐼𝐼 =1
12𝑀𝑀 𝐿𝐿2
= �−𝐿𝐿/2
𝐿𝐿/2𝑟𝑟2 𝑟𝑟𝑚𝑚
𝑟𝑟𝑚𝑚𝑟𝑟𝑑𝑑
=𝑀𝑀𝐿𝐿
𝑀𝑀
= �−𝐿𝐿/2
𝐿𝐿/2𝑑𝑑2
𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
𝑟𝑟 = 𝑑𝑑𝑟𝑟
𝐿𝐿/2
𝑟𝑟𝑚𝑚 =𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
17Prof. Sergio B. MendesSummer 2018
Example 10.5, modified, solution “a”
𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚
𝐼𝐼 =13𝑀𝑀 𝐿𝐿2
= �−𝐿𝐿/2
𝐿𝐿/2𝑟𝑟2 𝑟𝑟𝑚𝑚
𝑟𝑟𝑚𝑚 =𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
= �−𝐿𝐿/2
𝐿𝐿/2𝑑𝑑 +
𝐿𝐿2
2 𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
𝑟𝑟 = 𝑑𝑑𝑀𝑀 𝑟𝑟 = 𝑑𝑑 +
𝐿𝐿2
𝑟𝑟
18Prof. Sergio B. MendesSummer 2018
Example 10.5, modified, solution “b”
𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚
𝐼𝐼 =13𝑀𝑀 𝐿𝐿2
= �0
𝐿𝐿𝑟𝑟2 𝑟𝑟𝑚𝑚
𝑟𝑟𝑚𝑚 =𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
= �0
𝐿𝐿𝑑𝑑2
𝑀𝑀𝐿𝐿𝑟𝑟𝑑𝑑
𝑟𝑟
𝑟𝑟𝑑𝑑
𝑟𝑟 = 𝑑𝑑
𝑑𝑑0
19Prof. Sergio B. MendesSummer 2018
Example 10.6
𝐼𝐼 = 𝑀𝑀 𝑅𝑅2
𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚 = �𝑅𝑅2 𝑟𝑟𝑚𝑚 = 𝑅𝑅2 �𝑟𝑟𝑚𝑚
𝑀𝑀
20Prof. Sergio B. MendesSummer 2018
Example 10.7
𝐼𝐼 ≡ �𝑟𝑟2 𝑟𝑟𝑚𝑚
𝑟𝑟𝑚𝑚𝑟𝑟𝑑𝑑
=𝑀𝑀𝜋𝜋 𝑅𝑅2
𝑟𝑟𝑑𝑑 = 2 𝜋𝜋 𝑟𝑟 𝑟𝑟𝑟𝑟
𝑟𝑟𝑚𝑚 =𝑀𝑀𝜋𝜋 𝑅𝑅2
2 𝜋𝜋 𝑟𝑟 𝑟𝑟𝑟𝑟
= �0
𝑅𝑅𝑟𝑟2
𝑀𝑀𝜋𝜋 𝑅𝑅2
2 𝜋𝜋 𝑟𝑟 𝑟𝑟𝑟𝑟
𝐼𝐼 =12𝑀𝑀 𝑅𝑅2
𝑀𝑀
𝑟𝑟𝑚𝑚 =𝑀𝑀𝜋𝜋 𝑅𝑅2
𝑟𝑟𝑑𝑑
21Prof. Sergio B. MendesSummer 2018
Moment of Inertia: Different Shapes and Axis of Rotation
22Prof. Sergio B. MendesSummer 2018
Parallel Axis Theorem
𝐼𝐼 = ? ?
= 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝑀𝑀 𝑟𝑟2
Proof: Problem 10.78
23Prof. Sergio B. MendesSummer 2018
Now that we have learned how to calculate the moment of inertia of an object at an specific axis of rotation,
let’s apply it.
𝜏𝜏 = 𝐼𝐼 𝛼𝛼
24Prof. Sergio B. MendesSummer 2018
Example 10.8
2 𝑅𝑅 = 1.4 𝑚𝑚
𝑀𝑀 = 940 𝑘𝑘𝑘𝑘
𝜔𝜔𝑜𝑜 = 10 𝑟𝑟𝑟𝑟𝑚𝑚
𝐹𝐹 = 20 𝑁𝑁
𝜔𝜔 = 𝜔𝜔𝑜𝑜 − 𝛼𝛼 ∆𝑟𝑟
∆𝑟𝑟 = ? ?𝜔𝜔 = 0
∆𝑟𝑟 =𝜔𝜔 − 𝜔𝜔𝑜𝑜− 𝛼𝛼
𝜏𝜏 = 𝐼𝐼 𝛼𝛼 𝛼𝛼 =𝜏𝜏𝐼𝐼
𝜏𝜏 = 2 𝐹𝐹 𝑅𝑅 sin(90°)
𝐼𝐼 =12𝑀𝑀 𝑅𝑅2
=𝜔𝜔 −𝜔𝜔𝑜𝑜
−4 𝐹𝐹/ 𝑀𝑀𝑅𝑅
𝛼𝛼 = 𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
25Prof. Sergio B. MendesSummer 2018
Example 10.9
𝑟𝑟 = ? ?
𝑚𝑚𝑘𝑘 − 𝑇𝑇 = 𝑚𝑚 𝑟𝑟
𝜏𝜏 = 𝑇𝑇 𝑅𝑅 = 𝐼𝐼 𝛼𝛼𝑟𝑟 = 𝛼𝛼 𝑅𝑅
𝑅𝑅𝑀𝑀
𝑇𝑇 =𝐼𝐼 𝛼𝛼𝑅𝑅
=𝐼𝐼 𝑟𝑟𝑅𝑅2
𝐼𝐼 =12𝑀𝑀 𝑅𝑅2
𝑟𝑟 =𝑚𝑚 𝑘𝑘
𝑚𝑚 + 12𝑀𝑀
𝑀𝑀
𝑅𝑅
=𝑀𝑀 𝑟𝑟
2
26Prof. Sergio B. MendesSummer 2018
Kinetic Energy of a Rotating Object:
𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 = �12𝑟𝑟𝑚𝑚 𝑟𝑟2
= �12𝑟𝑟𝑚𝑚 𝜔𝜔 𝑟𝑟 2
=12𝜔𝜔2 �𝑟𝑟𝑚𝑚 𝑟𝑟2
=12𝜔𝜔2 𝐼𝐼 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 =
12𝐼𝐼 𝜔𝜔2
27Prof. Sergio B. MendesSummer 2018
Example 10.10
𝜔𝜔 = 31,000 𝑟𝑟𝑟𝑟𝑚𝑚
𝑅𝑅 = 30 𝑠𝑠𝑚𝑚
𝑀𝑀 = 135 𝑘𝑘𝑘𝑘Cylindrical Rotor
𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 =12𝐼𝐼 𝜔𝜔2
𝐼𝐼 =12𝑀𝑀 𝑅𝑅2
𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 = 32 𝑀𝑀𝑀𝑀
28Prof. Sergio B. MendesSummer 2018
Work and Kinetic Energy of a Rotating Object:
𝑊𝑊𝑟𝑟𝑜𝑜𝑡𝑡 = �𝐹𝐹⊥ 𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟 = 𝑟𝑟 𝑟𝑟𝜃𝜃= �𝐹𝐹⊥ 𝑟𝑟 𝑟𝑟𝜃𝜃
= �𝜏𝜏 𝑟𝑟𝜃𝜃𝐹𝐹⊥ 𝑟𝑟 = 𝜏𝜏
𝑟𝑟𝜃𝜃 = 𝜔𝜔 𝑟𝑟𝑟𝑟
𝜏𝜏 = 𝐼𝐼𝑟𝑟𝜔𝜔𝑟𝑟𝑟𝑟
= �𝐼𝐼𝑟𝑟𝜔𝜔𝑟𝑟𝑟𝑟
𝜔𝜔 𝑟𝑟𝑟𝑟
= �𝐼𝐼 𝜔𝜔 𝑟𝑟𝜔𝜔
𝑊𝑊𝑟𝑟𝑜𝑜𝑡𝑡 =12𝐼𝐼 𝜔𝜔𝑓𝑓2 −
12𝐼𝐼 𝜔𝜔𝑖𝑖
2
= 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡,𝑓𝑓 − 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡,𝑖𝑖
= ∆𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡
29Prof. Sergio B. MendesSummer 2018
𝑊𝑊𝑟𝑟𝑜𝑜𝑡𝑡 = �𝐹𝐹⊥ 𝑟𝑟𝑟𝑟 =12𝐼𝐼 𝜔𝜔𝑓𝑓2 −
12𝐼𝐼 𝜔𝜔𝑖𝑖
2 = ∆𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡
30Prof. Sergio B. MendesSummer 2018
Example 10.11
𝐼𝐼 = 2.7 𝑘𝑘𝑘𝑘 𝑚𝑚2
𝜏𝜏 = ? ?
𝑊𝑊𝑟𝑟𝑜𝑜𝑡𝑡 = �𝜃𝜃𝑖𝑖
𝜃𝜃𝑓𝑓𝜏𝜏 𝑟𝑟𝜃𝜃
𝜔𝜔𝑓𝑓 = 700 𝑟𝑟𝑟𝑟𝑚𝑚
𝜔𝜔𝑖𝑖 = 0 𝑟𝑟𝑟𝑟𝑚𝑚
∆𝜃𝜃 = 25 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
=12𝐼𝐼 𝜔𝜔𝑓𝑓2 −
12𝐼𝐼 𝜔𝜔𝑖𝑖
2
𝜏𝜏 �𝜃𝜃𝑖𝑖
𝜃𝜃𝑓𝑓𝑟𝑟𝜃𝜃 =
12𝐼𝐼 𝜔𝜔𝑓𝑓2𝜏𝜏 𝜃𝜃𝑓𝑓 − 𝜃𝜃𝑖𝑖 =𝜏𝜏 ∆𝜃𝜃 = 𝜏𝜏 =
𝐼𝐼 𝜔𝜔𝑓𝑓2
2 ∆𝜃𝜃
= 46 𝑁𝑁 𝑚𝑚
31Prof. Sergio B. MendesSummer 2018
Total Kinetic Energy
𝐾𝐾 = 𝐾𝐾𝐶𝐶𝐶𝐶 + �𝑖𝑖=1
𝑁𝑁
𝐾𝐾𝑖𝑖,𝑟𝑟𝑟𝑟𝑟𝑟
𝐾𝐾 = 𝐾𝐾𝐶𝐶𝐶𝐶 + 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡
for objects under rotational and
translational motions
32Prof. Sergio B. MendesSummer 2018
Rolling Objectsrolling without slipping
𝑟𝑟𝐶𝐶𝐶𝐶 =𝜋𝜋 𝑅𝑅∆𝑟𝑟
𝜔𝜔 =𝜋𝜋∆𝑟𝑟
𝑟𝑟𝑟𝑟𝑜𝑜𝑡𝑡 = 𝜔𝜔 𝑅𝑅
=𝜋𝜋 𝑅𝑅∆𝑟𝑟
= 𝑟𝑟𝐶𝐶𝐶𝐶
∆𝑟𝑟
33Prof. Sergio B. MendesSummer 2018
Work done by Friction:
𝑊𝑊𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓 = �𝑥𝑥𝑖𝑖
𝑥𝑥𝑓𝑓𝐹𝐹𝑥𝑥 𝑟𝑟𝑑𝑑
𝑊𝑊𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓 = 0
𝑊𝑊𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓 = 0
𝑊𝑊𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓 < 0
rolling without slipping
𝑟𝑟
𝑟𝑟
𝐹𝐹𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝐹𝐹𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝐹𝐹𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓
𝐹𝐹𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓
static friction
kinetic friction
static friction
𝑟𝑟 = 0
𝑟𝑟 = 0
34Prof. Sergio B. MendesSummer 2018
𝑊𝑊𝑓𝑓𝑟𝑟𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑜𝑜𝑓𝑓 = 0
𝐾𝐾 + 𝑈𝑈 = 𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
Conservation of Mechanical Energy for a rolling object:
rolling without slipping
𝐾𝐾 = 𝐾𝐾𝐶𝐶𝐶𝐶 + 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡
35Prof. Sergio B. MendesSummer 2018
Example 10.12𝑀𝑀
𝑅𝑅 𝑟𝑟 = ? ?
𝐾𝐾 = 𝐾𝐾𝐶𝐶𝐶𝐶 + 𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡
𝑈𝑈 = 𝑀𝑀 𝑘𝑘 𝑦𝑦
𝐾𝐾 + 𝑈𝑈 = 𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝐾𝐾𝐶𝐶𝐶𝐶 =12𝑀𝑀 𝑟𝑟𝐶𝐶𝐶𝐶2
𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 =12 𝐼𝐼 𝜔𝜔
2 =12 𝐼𝐼
𝑟𝑟𝐶𝐶𝐶𝐶𝑅𝑅
2
𝑈𝑈 = 𝑀𝑀 𝑘𝑘 ℎ
𝐾𝐾𝐶𝐶𝐶𝐶 =12𝑀𝑀 𝑟𝑟𝐶𝐶𝐶𝐶2
𝐾𝐾𝑟𝑟𝑜𝑜𝑡𝑡 =12𝐼𝐼𝑟𝑟𝐶𝐶𝐶𝐶𝑅𝑅
2
𝑟𝑟𝐶𝐶𝐶𝐶 =2 𝑘𝑘 ℎ
1 + 𝐼𝐼𝑀𝑀 𝑅𝑅2
𝐼𝐼 =25
M 𝑅𝑅2
=10 𝑘𝑘 ℎ
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