Physics 111. Lecture 24 (Walker:11.2-3) Rotational Dynamics Equilibrium of Solid Objects April 1, Gravitational Torque

Transcription

1 Physics 111 Lecture 24 (Walker:11.2-3) Rotational Dynamics Equilibrium of Solid Objects April 1, 2009 Lecture 24 1/24 Gravitational Torque An object fixed on a pivot (taken as the origin) will experience gravitational forces that will produce torques. The net torque due to gravity acts as if all mass is at the CM. The minus sign is because a CM to the right of the origin (x positive) will produce clockwise (negative) torques. τ = Mg x grav cm Lecture 24 2/24

2 Ex.: Gravitational Torque on Beam A 4.0 m long 500 kg steel beam is supported 1.20 m from the right end. What is the gravitational torque about the support? τ = = = 2 grav Mgxcm (500 kg)(9.80 m/s )( 0.80m) 3920 Nm Lecture 24 3/24 Example: Determine the net torque: N 4 m 2 m 500 N 800 N 1. Draw all applicable forces 2. Consider CCW rotation to be positive τ = (500 N)(4 m) + ( )(800 N)(2 m) = N m 1600 N m =+ 400 N m Rotation would be CCW Lecture 24 4/24

3 Torque & Angular Acceleration Newton s Second Law: If we consider a mass m rotating around an axis a distance r away, we can reformat Newton s Second Law to read: Or equivalently, Lecture 24 5/24 Torque & Angular Acceleration Once again, we have analogies between linear and angular motion: F = m a τ = I α Lecture 24 6/24

4 Problem Solving Strategy Picture: Angular acceleration for rigid objects can be found using Newton s 2 nd Law for rotation, τ net ext = Στ ext = Iα. If τ net ext is constant, then constant angular acceleration equations apply. Times, angular positions, angular velocities, and angular accelerations can be found. Solve: 1. Draw a free-body diagram with the object shown as a likeness of the object (not as a dot). 2. Draw each force vector along the line of action of that force and find lever arm for each force. 3. On diagram, show positive direction (cw or ccw) for rotations. Check: Make sure that signs of results are consistent with choice of positive directions of rotation. Lecture 24 7/24 Example: Stationary Bike As you pedal, chain applies a force of F = 18 N to rear sprocket wheel at a distance of r s = 7.0 cm from rotation axis of the wheel. Consider the wheel to be a hoop of radius R = 35 cm and mass M = 2.4 kg. What is the angular velocity ω of the wheel after 5.0 s? Lecture 24 8/24

5 τ Stationary Bike net = Fr = τ I net α = = s F r s MR Iα ω = ω + α = + αt 2 0 t 0 Fr (18 N)(0.070 m) ω= αt= s t= 2 2 (5.0 s) = 21.4 rad/s MR (2.4 kg)(0.35 m) Lecture 24 9/24 Example:Uniform Rod Pivoted at End A uniform thin rod of length L and mass M is pivoted at one end. It is held horizontal and released. Neglect friction and air drag. Find angular acceleration α of rod immediately after its release. τ ext = Iα grav 2 1 ( ) τ = Mg L 1 ( 2 ) α = grav = 1 2 = 3 I = τ Mg L 3g I ML 2L ML Lecture 24 10/24

6 Static Equilibrium An object with forces acting on it, but that is not moving (or rotating), is said to be in equilibrium. Lecture 24 11/24 Conditions for Equilibrium The first condition for equilibrium is that the forces along each coordinate axis add to zero. Lecture 24 12/24

7 Conditions for Equilibrium The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary. Lecture 24 13/24 Zero Torque & Static Equilibrium Static equilibrium occurs when an object is at rest, neither rotating nor translating. Lecture 24 14/24

8 Zero Torque & Static Equilibrium If net torque is zero, doesn t matter which axis we consider rotation to be around; can choose the one that makes our calculations easiest. Often choose axis through which unknown force acts. Axis Axis Lecture 24 15/24 Solving Statics Problems If a force in your solution comes out negative (as F A will here), it just means that the force is actually in the opposite direction from the one you chose. Lecture 24 16/24

9 ConcepTest A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? Balancing Rod 1m 1) 1/4 kg 2) 1/2 kg 3) 1 kg 4) 2 kg 5) 4 kg 1kg Example 2: What distance d 2 should girl have from axis for equilb m? What will be normal force N? (Neglect see-saw mass.) τ RHS = (800 N)(2 m) d 2 2 m N y 500 N 800 N 1. Draw all applicable forces and lever arms τ LHS = (500 N)( d2 m) [ N m] d [ N m] = 0 d = 3.2 m F i = ( 500 N) + N' + ( 800 N) = 0 N' = 1300 N 2 2 Lecture 24 18/24

10 Example: Ladder 10 m ladder weighing 50N placed against smooth wall at 50 angle. Find forces from floor & wall, and the µ s needed for equilibrium? 1. Draw all applicable forces 2. Choose axis of rotation at bottom corner (τ of f and n are 0!) Lecture 24 19/24 Torque Condition: L τ = mg cos50 PLsin 50 = 0 2 P sin 50 = (25N)cos50 P = 21N Force Conditions: f F = P = 21N F x y = 0 = f P = 0 = n mg n = mg = 50N f µ s n so need µ s (f/n) or µ s 0.42 Lecture 24 20/24

11 Solving Statics Problems If there is a cable or cord (or muscle) in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend. Lecture 24 21/24 Applications to Muscles and Joints These same principles can be used to understand forces within the body. Lecture 24 22/24

12 Force to Hold Baseball (Prob. 11-5) Person holds 1.4N baseball in hand, 0.34 m from elbow joint. Biceps, attached at 2.75 cm from elbow, exerts 12.6 N force. Take forearm & hand to be uniform rod of mass 1.2 kg. Find net torque on forearm & hand. Take axis to be elbow joint. Στ=(0.0275m)(12.6N) -(0.17m)(1.2kg)(9.8N/kg) -(0.34m)(1.4N)=-2.1 Nm Lecture 24 23/24 Applications to Muscles and Joints The angle at which this man s back is bent places an enormous force on the disks at the base of his spine, as the lever arm for F M is so small. Lecture 24 24/24

13 Center of Mass & Balance If an extended object is to be balanced, it must be supported through its center of mass. Lecture 24 25/24 Balance An object balances when the pivot point is located directly under (or over) the center of mass. For that situation, τ grav =0. Lecture 24 26/24

14 Center of Mass & Balance This fact can be used to find the center of mass of an object suspend it from different axes and trace a vertical line. The center of mass is where the lines meet. Lecture 24 27/24 End of Lecture 24 For Friday, read Walker Homework Assignment #11a is due at 11:00 PM on Friday, April 3. Lecture 24 28/24