Momentum. The way to catch a knuckleball is to wait until it stops rolling and then pick it up. -Bob Uecker
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1 Chapter 11 -, Chapter 11 -, Angular The way to catch a knuckleball is to wait until it stops rolling and then pick it up. -Bob Uecker David J. Starling Penn State Hazleton PHYS 211
2 Chapter 11 -, motion is a combination of pure rotation and pure translation. Angular
3 Chapter 11 -, motion is a combination of pure rotation and pure translation. Angular
4 Chapter 11 -, motion is a combination of pure rotation and pure translation. Angular
5 Chapter 11 -, motion is a combination of pure rotation and pure translation. Angular Each velocity vector on the wheel is a sum of v com and the rotation.
6 Chapter 11 -, The relationship between the angular and linear velocities is fixed if the wheel does not slip. v com = Rω Angular
7 Chapter 11 -, The relationship between the angular and linear velocities is fixed if the wheel does not slip. v com = Rω Angular Note: therefore, a com = Rα.
8 Chapter 11 -, If the rolling object does not slip, then the point in contact with the floor is momentarily stationary. Angular
9 Chapter 11 -, If the rolling object does not slip, then the point in contact with the floor is momentarily stationary. Angular In this case, there is a static force of friction at the pivot that does no work.
10 Chapter 11 -, The kinetic energy of the wheel rotating about point P is Angular K = 1 2 I pω 2
11 Chapter 11 -, The kinetic energy of the wheel rotating about point P is Angular K = 1 2 I pω 2 = 1 ( Icom + MR 2) ω 2 2
12 Chapter 11 -, The kinetic energy of the wheel rotating about point P is Angular K = 1 2 I pω 2 = 1 ( Icom + MR 2) ω 2 2 = 1 2 I comω M(Rω)2
13 Chapter 11 -, The kinetic energy of the wheel rotating about point P is Angular K = 1 2 I pω 2 = 1 ( Icom + MR 2) ω 2 2 = 1 2 I comω M(Rω)2 = 1 2 I comω Mv2
14 Chapter 11 -, The kinetic energy of the wheel rotating about point P is Angular K = 1 2 I pω 2 = 1 ( Icom + MR 2) ω 2 2 = 1 2 I comω M(Rω)2 = 1 2 I comω Mv2 K = K R + K T
15 Chapter 11 -, When a rolling object accelerates, the friction force opposes the tendency to slip. Angular
16 Chapter 11 -, When a rolling object accelerates, the friction force opposes the tendency to slip. Angular Notice how the force does not oppose the direction of motion!
17 Chapter 11 -, Lecture Question 11.1 Which one of the following statements concerning a wheel undergoing rolling motion is true? (a) The angular acceleration of the wheel must be 0 m/s 2. (b) The tangential velocity is the same for all points on the wheel. (c) The linear velocity for all points on the rim of the wheel is non-zero. (d) The tangential velocity is the same for all points on the rim of the wheel. (e) There is no slipping at the point where the wheel touches the surface on which it is rolling. Angular
18 Angular Chapter 11 -, An object A with momentum p also has angular momentum l = r p about some point O. Angular
19 Angular Chapter 11 -, An object A with momentum p also has angular momentum l = r p about some point O. Angular Why is it defined this way?
20 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt
21 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt = d r d p p + r dt dt
22 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt = d r d p p + r dt dt = v p + r (m a)
23 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt = d r d p p + r dt dt = v p + r (m a) = r F net
24 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt = d r d p p + r dt dt = v p + r (m a) = r F net = τ net
25 Angular Chapter 11 -, From Newton s 2nd Law, let s take a derivative of this momentum : Angular d l dt = d ( r p) dt = d r d p p + r dt dt = v p + r (m a) = r F net = τ net τ net = d l dt
26 Angular Chapter 11 -, If there is more than one particle, the total angular momentum L is just the vector sum of the individual angular momentums l i. Angular L = l 1 + l l N = N l i i=1 And the net torque on the whole system is just: τ net = d L dt
27 Angular The angular momentum of a rigid body can be found by adding up the angular momentum of the particles that make it up: Chapter 11 -, Angular l i = r i p i = r i m i v i
28 Angular Chapter 11 -, Angular l i = r i m i v i
29 Angular Chapter 11 -, Angular l i = r i m i v i = r i m i (r i ω)
30 Angular Chapter 11 -, Angular l i = r i m i v i = r i m i (r i ω) = ( m i r 2 i )ω
31 Angular Chapter 11 -, Angular l i = r i m i v i = r i m i (r i ω) = ( m i r 2 i )ω L = i ( m i r 2 i )ω
32 Angular Chapter 11 -, Angular l i = r i m i v i = r i m i (r i ω) = ( m i r 2 i )ω L = ( m i r i 2 )ω i ( ) = m i r i 2 ω i
33 Angular Chapter 11 -, Angular l i = r i m i v i = r i m i (r i ω) = ( m i r 2 i )ω L = ( m i r i 2 )ω i ( ) = m i r i 2 ω L = I ω i
34 Angular Chapter 11 -, There are many similarities between linear and angular momentum. Angular
35 Angular Chapter 11 -, Lecture Question 11.2 What is the direction of the Earth s angular momentum as it spins about its axis? (a) north (b) south (c) east (d) west (e) radially inward Angular
36 If the net external torque on a system is zero, the angular momentum is conserved. τ net = d L dt = 0 L = constant Chapter 11 -, Angular
37 If the net external torque on a system is zero, the angular momentum is conserved. τ net = d L dt = 0 L = constant Chapter 11 -, Angular
38 A gyroscope is an object that spins very quickly. Such an object behaves obeys the angular Newton s Second Law. Chapter 11 -, Angular τ = d L dt
39 A gyroscope is an object that spins very quickly. Such an object behaves obeys the angular Newton s Second Law. Chapter 11 -, Angular τ = d L dt
40 A gyroscope is an object that spins very quickly. Such an object behaves obeys the angular Newton s Second Law. Chapter 11 -, Angular τ = d L dt
41 When a gyroscope is subjected to a gravitational force it precesses. Chapter 11 -, Angular
42 When a gyroscope is subjected to a gravitational force it precesses. Chapter 11 -, Angular τ = d L dt
43 When a gyroscope is subjected to a gravitational force it precesses. Chapter 11 -, Angular τ = d L dt dl = τ dt = Mgr dt
44 When a gyroscope is subjected to a gravitational force it precesses. Chapter 11 -, Angular τ = d L dt dl = τ dt = Mgr dt dφ = dl Mgr dt = L Iω
45 When a gyroscope is subjected to a gravitational force it precesses. Chapter 11 -, Angular τ = d L dt dl = τ dt = Mgr dt dφ = dl L dφ dt = Mgr Iω = Mgr dt Iω (rate of precession)
46 Chapter 11 -, Lecture Question 11.3 A solid sphere of radius R rotates about an axis that is tangent to the sphere with an angular speed ω. Under the action of internal forces, the radius of the sphere increases to 2R. What is the final angular speed of the sphere? (a) ω/4 (b) ω/2 (c) ω (d) 2ω (e) 4ω Angular
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