Download presentation

Presentation is loading. Please wait.

Published byAden Haymaker Modified over 3 years ago

1
Physics 111: Lecture 19, Pg 1 Physics 111: Lecture 19 Today’s Agenda l Review l Many body dynamics l Weight and massive pulley l Rolling and sliding examples l Rotation around a moving axis: Puck on ice l Rolling down an incline l Bowling ball: sliding to rolling l Atwood’s Machine with a massive pulley

2
Physics 111: Lecture 19, Pg 2 Review: Direction & The Right Hand Rule l To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector! l We normally pick the z-axis to be the rotation axis as shown. = z = z = z l For simplicity we omit the subscripts unless explicitly needed. x y z x y z

3
Physics 111: Lecture 19, Pg 3 Review: Torque and Angular Acceleration NET = I l This is the rotational analogue of F NET = ma l Torque is the rotational analogue of force: çThe amount of “twist” provided by a force. Moment of inertia I is the rotational analogue of mass Moment of inertia I is the rotational analogue of mass If I is big, more torque is required to achieve a given angular acceleration.

4
Physics 111: Lecture 19, Pg 4 Lecture 19, Act 1 Rotations l Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. l Forces F 1 and F 2 are applied as shown. What is F 2 / F 1 if the angular acceleration of the wheels is the same? (a) 1 (b) 2 (c) 4 F1F1 F2F2

5
Physics 111: Lecture 19, Pg 5 Lecture 19, Act 1 Solution We know but and so F1F1 F2F2 Since R 2 = 2 R 1

6
Physics 111: Lecture 19, Pg 6 Review: Work & Energy The work done by a torque acting through a displacement is given by: l The power provided by a constant torque is therefore given by:

7
Physics 111: Lecture 19, Pg 7 Falling weight & pulley A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley. çStarting at rest, how long does it take for the mass to fall a distance L. I m R T mg a L

8
Physics 111: Lecture 19, Pg 8 Falling weight & pulley... l For the hanging mass use F = ma çmg - T = ma For the pulley + flywheel use = I = TR = I Realize that a = R l Now solve for a using the above equations. I m R T mg a L

9
Physics 111: Lecture 19, Pg 9 Falling weight & pulley... l Using 1-D kinematics (Lecture 1) we can solve for the time required for the weight to fall a distance L: I m R T mg a L where Flywheel w/ weight

10
Physics 111: Lecture 19, Pg 10 Rotation around a moving axis. l A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds. çWhat length of string L has unwound after the puck has moved a distance D? F R M Top view

11
Physics 111: Lecture 19, Pg 11 Rotation around a moving axis... l The CM moves according to F = MA F MA l The distance moved by the CM is thus R The disk will rotate about its CM according to = I l So the angular displacement is

12
Physics 111: Lecture 19, Pg 12 Rotation around a moving axis... So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time: D F F Divide (b) by (a): (a)(b) L The length of string pulled out is L = R :

13
Physics 111: Lecture 19, Pg 13 Comments on CM acceleration: We just used = I for rotation about an axis through the CM even though the CM was accelerating! çThe CM is not an inertial reference frame! Is this OK?? (After all, we can only use F = ma in an inertial reference frame). YES! YES! We can always write = I for an axis through the CM. çThis is true even if the CM is accelerating. çWe will prove this when we discuss angular momentum! F R MA

14
Physics 111: Lecture 19, Pg 14 Rolling An object with mass M, radius R, and moment of inertia I rolls without slipping down a plane inclined at an angle with respect to horizontal. What is its acceleration? l Consider CM motion and rotation about the CM separately when solving this problem (like we did with the last problem)... R I M

15
Physics 111: Lecture 19, Pg 15 Rolling... l Static friction f causes rolling. It is an unknown, so we must solve for it. l First consider the free body diagram of the object and use F NET = MA CM : In the x direction Mg sin - f = MA Now consider rotation about the CM and use = I realizing that = Rf and A = R R M f Mg y x

16
Physics 111: Lecture 19, Pg 16 Rolling... l We have two equations: l We can combine these to eliminate f: A R I M For a sphere:

17
Physics 111: Lecture 19, Pg 17 Lecture 19, Act 2 Rotations l Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other. çIf both are placed at the top of the same ramp and released, which is moving faster at the bottom? (a) bigger one (b) smaller one (c) same

18
Physics 111: Lecture 19, Pg 18 Lecture 19, Act 2 Solution l Consider one of them. Say it has radius R, mass M and falls a height H. H Energy conservation: - U = K butand

19
Physics 111: Lecture 19, Pg 19 Lecture 19, Act 2 Solution H So: So, (c) does not depend on size, as long as the shape is the same!!

20
Physics 111: Lecture 19, Pg 20 Sliding to Rolling A bowling ball of mass M and radius R is thrown with initial velocity v 0. It is initially not rotating. After sliding with kinetic friction along the lane for a distance D it finally rolls without slipping and has a new velocity v f. The coefficient of kinetic friction between the ball and the lane is . çWhat is the final velocity, v f, of the ball? v f = R f = Mg v0v0 D Roll bowling ball

21
Physics 111: Lecture 19, Pg 21 Sliding to Rolling... While sliding, the force of friction will accelerate the ball in the -x direction: F = - Mg = Ma so a = - g The speed of the ball is therefore v = v 0 - gt (a) Friction also provides a torque about the CM of the ball. Using = I and remembering that I = 2 / 5 MR 2 for a solid sphere about an axis through its CM: D x f = Mg (b) v f = R v0v0

22
Physics 111: Lecture 19, Pg 22 Sliding to Rolling... l We have two equations: Using (b) we can solve for t as a function of Plugging this into (a) and using v f = R (the condition for rolling without slipping): D x (a)(b) f = Mg Doesn’t depend on , M, g !! v f = R v0v0

23
Physics 111: Lecture 19, Pg 23 Lecture 19, Act 3 Rotations l A bowling ball (uniform solid sphere) rolls along the floor without slipping. çWhat is the ratio of its rotational kinetic energy to its translational kinetic energy? Recall that for a solid sphere about an axis through its CM: (a) (b) (c)

24
Physics 111: Lecture 19, Pg 24 Lecture 19, Act 3 Solution l The total kinetic energy is partly due to rotation and partly due to translation (CM motion). rotational K translational K K =

25
Physics 111: Lecture 19, Pg 25 Lecture 19, Act 3 Solution Since it rolls without slipping: rotational K Translational K K =

26
Physics 111: Lecture 19, Pg 26 Atwoods Machine with Massive Pulley: l A pair of masses are hung over a massive disk-shaped pulley as shown. çFind the acceleration of the blocks. m2m2 m1m1 R M y x m2gm2g a T1T1 m1gm1g a T2T2 l For the hanging masses use F = ma ç -m 1 g + T 1 = -m 1 a ç -m 2 g + T 2 = m 2 a (Since for a disk) For the pulley use = I ç T 1 R - T 2 R

27
Physics 111: Lecture 19, Pg 27 Atwoods Machine with Massive Pulley... l We have three equations and three unknowns (T 1, T 2, a). Solve for a. -m 1 g + T 1 = -m 1 a (1) -m 2 g + T 2 = m 2 a (2) T 1 - T 2 (3) Large and small pulleys m2m2 m1m1 R M y x m2m2 m1m1 m2gm2g a T1T1 m1gm1g a T2T2

28
Physics 111: Lecture 19, Pg 28 Recap of today’s lecture l Review (Text: 9-1 to 9-6) l Many body dynamics l Weight and massive pulley (Text: 9-4) l Rolling and sliding examples (Text: 9-6) l Rotation around a moving axis: Puck on ice (Text: 9-4) l Rolling down an incline (Text: 9-6) l Bowling ball: sliding to rolling l Atwood’s Machine with a massive pulley (Text: 9-4) l Look at textbook problems l Look at textbook problems Chapter 9: # 53, 89, 92, 113, 125

Similar presentations

OK

Chapter 11 Angular Momentum. The Vector Product and Torque The torque vector lies in a direction perpendicular to the plane formed by the position vector.

Chapter 11 Angular Momentum. The Vector Product and Torque The torque vector lies in a direction perpendicular to the plane formed by the position vector.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Evs ppt on pollution Ppt on brushless dc motor Ppt on the great revolt of 1857 Ppt on trial and error learning Ppt on security guard training Ppt on cash reserve ratio Ppt on pronouns for grade 3 Ppt on religious tourism in india Plant life cycle for kids ppt on batteries Ppt on cartesian product multiplication