Physics 106P: Lecture 1 Notes

Physics 106P: Lecture 1 Notes

Exam II Physics 101: Lecture 15 Rolling Objects Todays lecture will cover Textbook Chapter 8.58.7 Physics 101: Lecture 15, Pg 1 Overview Review Krotation = I 2 Torque = Force that causes rotation = F r sin Equilibrium F = 0 = 0 Today = I rotational F = ma) nergy conservation revisited Physics 101: Lecture 15, Pg 2 Linear and Angular Displacement Velocity Acceleration Inertia KE N2L

Momentum Linear x v a m m v2 F=ma p = mv Angular I I 2 = I L = I Toda y Physics 101: Lecture 15, Pg 3 Rotational Form Newtons 2 Law nd

=I Torque is amount of twist provide by a force Signs: positive = CCW Moment of Inertia like mass. Large I means hard to start or stop from spinning. Problems Solved Like N2L Draw FBD Write N2L Physics 101: Lecture 15, Pg 4 The Hammer! You want to balance a hammer on the tip of your finger, which way is easier 38% A) Head up 58% B) Head down 4% C) Same Why am I balancing a hammer on my finger? It sounds dangerous.

I just tried it in my home and I guess it is easier to balance the hammer with the head up. Angular acceleration is smaller the larger the radius the larger the moment of inertia. Physics 101: Lecture 15, Pg 5 The Hammer! You want to balance a hammer on the tip of your finger, which way is easier 38% A) Head up 58% B) Head down R 4% C) Same =I Key idea: higher angular m g R sin() = mR2 means more acceleration Angular acceleration Torque Inertia difficult to balance. increases decreases with R! increase with R s as R2 So large R is easier to balance. g sin() / R =

What is angular acceleration? m g Physics 101: Lecture 15, Pg 6 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. Starting at rest, how long does it take for the mass to fall a distance L. I R T m What method should we use to solve this problem?

a mg L A) Conservation of Energy (including rotational) B)) and then use kinematics Either would work, but since it asks for time, we will use B. Physics 101: Lecture 15, Pg 7 Falling weight & pulley... For the hanging mass use F = ma mg - T = ma For the flywheel use = I I R TR sin(90) = I Realize

that a = R TR I Now T a R m solve for a, eliminate T: a mg L mR 2 a g 2 mR I Physics 101: Lecture 15, Pg 8 Falling weight & pulley... Using 1-D kinematics we

can solve for the time required for the weight to fall a distance L: R 1 2 y y 0 v0 t at 2 1 L at 2 2 where 2L t a mR 2 a g 2 mR I I T m a

mg L Physics 101: Lecture 15, Pg 9 Torque ACT Which pulley will make it drop fastest? 1) Small pulley 2) Large pulley 3) Same mR 2 a g 2 mR I Larger R, gives larger acceleration. Physics 101: Lecture 15, Pg 10 25 Tension m1 T1

m2 T2 F m3 Compare the tensions T1 and T2 as the blocks are accelerated to the right by the force F. A) T1 < T2 B) T1 = T2 C) T1 > T2 T1 < T2 since T2 T1 = m2 a. It takes force to accelerate block 2. T1 m2 Compare the tensions T1 and T2 as block 3 falls A) T1 < T2 B) T1 = T2 C) T1 > T2 m1 T2

m3 T2 > T1 since RT2 RT1 = I2 . It takes force (torque) to accelerate the pulley. Physics 101: Lecture 15, Pg 11 Rolling y A wheel is spinning clockwise such that the speed of the outer rim is 2 m/s. What is the velocity of the top of the wheel relative to the ground? + 2 m/s What is the velocity of the bottom of the wheel relative to the ground? -2 m/s x 2 m/s 2 m/s You now carry the spinning wheel to the right at 2 m/s. What is the velocity of the top of the wheel relative to the ground? A) -4 m/s B) -2 m/s C) 0 m/s

D) +2m/s E) +4 m/s What is the velocity of the bottom of the wheel relative to the ground? A) -4 m/s B) -2 m/s C) 0 m/s D) +2m/s E) +4 m/s Physics 101: Lecture 15, Pg 12 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? Consider CM motion and rotation about the CM separately when solving this problem

I M R Physics 101: Lecture 15, Pg 13 Rolling... Static friction f causes rolling. It is an unknown, so we must solve for it. First consider the free body diagram of the object and use FNET = Macm : In the x direction Mg sin - f = Macm M Now consider rotation about the CM y and use = I realizing that = Rf and a = R Rf I a R

f I a R2 R x Mg Physics 101: Lecture 15, Pg 14 f Rolling... We have two equations: Mg sin - f = Ma a f I 2 R We can combine these to eliminate f:

MR 2sin a g 2 MR I I A M R For a sphere: 2 MR sin 5 a g 7 gsin 2 MR 2 MR 2 5

Physics 101: Lecture 15, Pg 15 Energy Conservation! Friction causes object to roll, but if it rolls w/ o slipping friction does NO work! W = F d cos d is zero for point in contact No dissipated work, energy is conserved Need to include both translational and rotational kinetic energy. K = m v2 + I 2 Physics 101: Lecture 15, Pg 16 Translational + Rotational KE Consider a cylinder with radius R and mass M, rolling w/o slipping down a ramp. Determine the ratio of the translational to rotational KE.

Translational: KT = M v2 Rotational: KR = I 2 use 1 I MR 2 and 2 Rotational: V R KR = ( M R2) V/R)2 = M v2 H = KT Physics 101: Lecture 15, Pg 17 Rolling Act

Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other. If 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 Ki + Ui = Kf + Uf 1 1 MgH I 2 MV 2 2 2 2 1 1 V 1 MgH MR 2 2 MV 2 R 22

2 V 4 gH 3 Physics 101: Lecture 15, Pg 18 Summary =I Energy is Conserved Need to include translational and rotational Physics 101: Lecture 15, Pg 19

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