PHYSICS 220 Lecture 14 Equilibrium and Dynamics Lecture

PHYSICS 220 Lecture 14 Equilibrium and Dynamics Lecture 14 Purdue University, Physics 220 1

Lecture 14 Purdue University, Physics 220 2

Find Center of Mass Not in equilibrium Equilibrium pivot d W=mg Center of mass Torque about pivot 0 Torque about pivot = 0 A method to find center of mass of an irregular object Lecture 14 Purdue University, Physics 220 3

Exercise A 50 kg diver stands at the end of a 4. 6 m diving board. Neglecting the weight of the board, what is the force on the pivot 1. 2 meters from the end? 1) Draw FBD 2) Choose Axis of rotation 3) = 0 Rotational EQ F 1 (1. 2) – mg (4. 6) = 0 F 1 = 4. 6 (50 *9. 8) / 1. 2 F 1 = 1880 N 4) F = 0 Translational EQ F 1 – F 2 – mg = 0 F 2 = F 1 – mg = 1390 N Lecture 14 Purdue University, Physics 220 F 1 F 2 mg 4

Exercise A 75 kg painter stands at the center of a 50 kg 3 meter plank. The supports are 1 meter in from each edge. Calculate the force on support A. 1 meter B A FA 1) Draw FBD 2) F = 0 1 meter FA + FB – mg – Mg = 0 3) Choose pivot FB 0. 5 meter mg Mg 4) = 0 -FA (1) sin(90)+ FB (0) sin(90) + mg (0. 5)sin(90) + Mg(0. 5) sin(90) = 0 FA = 0. 5 mg + 0. 5 Mg = 612. 5 N Lecture 14 Purdue University, Physics 220 5

i. Clicker If the painter moves to the right, the force exerted by support A A) Increases B) Unchanged C) Decreases 1 meter A Lecture 14 B Purdue University, Physics 220 6

Question How far to the right of support B can the painter stand before the plank tips? 1 meter B A Just before board tips, force from A becomes zero FB 1) Draw FBD 2) F = 0 FB – mg – Mg = 0 0. 5 meter 3) Choose pivot mg 4) = 0 x Mg FB (0) sin(90) + mg (0. 5)sin(90) – Mg(x) sin(90) = 0 0. 5 m = x M -> x = 0. 3 meter Lecture 14 Purdue University, Physics 220 7

Spool on a Rough Surface • Consider all of the forces acting: tension T and friction f. – Using FNET = 0 in the x direction: Using NET = 0 about the CM axis: T Solving: a y b x f Lecture 14 Purdue University, Physics 220 8

Spool on a Rough Surface • There is another (slick) way to see this: • Consider the torque about the point of contact between the spool and the ground. We know the net torque about this (or any other) point is zero. – Since both Mg and f act through this point, they do not contribute to the net toque. T – Therefore the torque due to T must also be zero. – Therefore T must act y a along a line that passes b Mg through this point! x f Lecture 14 Purdue University, Physics 220 9

Spool on a Rough Surface • So we can use geometry to get the same result T a b Lecture 14 Purdue University, Physics 220 10

Rotational Dynamics =I – Torque is amount of twist provide by a force • Signs: positive = CCW – Moment of Inertial like mass. Large I means hard to start or stop from spinning. – Angular acceleration is defined as the rate at which the angular velocity changes. • Problems Solving: – Draw free body diagram – Pick an axis – Compute torque due to each force Lecture 14 Purdue University, Physics 220 11

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 a mg L Lecture 14 Purdue University, Physics 220 12

Falling Weight & Pulley • Using 1 -D kinematics we can solve for the time required for the weight to fall a distance L: I R T m a mg L Lecture 14 Purdue University, Physics 220 13

Falling Weight & Pulley • For the hanging mass use F = ma mg - T = ma • For the flywheel use t = I I R = TR sin(90) = I • Realize that a = R • Now solve for a using the above equations. Lecture 14 Purdue University, Physics 220 T m a mg L 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? • Consider CM motion and rotation about the CM separately when solving this problem I M R Lecture 14 Purdue University, Physics 220 15

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 • Now consider rotation about the CM and use = I realizing that = Rf and acm = R Lecture 14 Purdue University, Physics 220 M y f R x Mg 16

Rolling • We have two equations: Mg sin - f = Ma • We can combine these to eliminate f: I For a sphere: a M R Lecture 14 Purdue University, Physics 220 17

Energy Conservation • Friction causes an object to roll, but if it rolls w/o slipping friction does NO work! • No dissipated work, energy is conserved • Need to include both translation and rotation kinetic energy. KE = ½ m vcm 2 + ½ I 2 Lecture 14 Purdue University, Physics 220 18

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: KET = ½ M Vcm 2 Rotational: KER = ½ I 2 use and KER = ½ (½ M R 2) (Vcm/R)2 = ¼ M Vcm 2 H Lecture 14 = ½ KET Purdue University, Physics 220 19

i. Clicker A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greater kinetic energy at bottom? A) Hoop B) Same C) Cylinder Lecture 14 Purdue University, Physics 220 20

Rolling Race (Hoop vs Cylinder) A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greater speed at the bottom of the ramp? A) Hoop B) Same C) Cylinder I = ½ MR 2 I = MR 2 Cylinder will get to the bottom first because inertia for a cylinder is less than that for a hoop type object. Lecture 14 Purdue University, Physics 220 21