Physics 111. Lecture 22 (Walker: ) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009

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Physics 111 Lecture 22 (Walker: 11.1-3) Torque Rotational Dynamics Static Equilibrium Oct.

For force F applied perpendicular distance r from the rotation axis: τ = r F The torque increases as the force

Lecture 22 2/26 Other Torque Units Common US torque unit: foot-pound (ft-lb) Seen in wheel lug nut torques for

1 Physics 111 Lecture 22 (Walker: ) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009 Lecture 22 1/26 Torque (τ) We define a quantity called torque which is a measure of twisting effort. For force F applied perpendicular distance r from the rotation axis: τ = r F The torque increases as the force increases and also as the perpendicular distance r (the lever arm ) increases. Lecture 22 2/26 Other Torque Units Common US torque unit: foot-pound (ft-lb) Seen in wheel lug nut torques for vehicles: Ford Aspire ft-lbs Ford Contour ft-lbs Ford Crown Victoria ft-lbs Longer lever arm is very helpful in rotating objects. Torque Amplifier Lecture 22 4/26 Lecture 22 3/26

Finding the Lever Arm r Here, the lever arm for F A is the distance from the knob to the hinge; the lever arm for F D is zero; and the lever arm for F C is as shown.

2 Finding the Lever Arm r Here, the lever arm for F A is the distance from the knob to the hinge; the lever arm for F D is zero; and the lever arm for F C is as shown. Lecture 22 5/26 Torque in Terms of Tangential Force The torque is defined as: Could also write this as τ = r F where r is the distance from the axis to where the force is applied and F is the tangential force component. Lecture 22 6/26 Torque This leads to a different definition of torque (as given in the textbook): where F is the force vector, r is the vector from the axis to the point where force is applied, and θ is the angle between r and F. (Angle measured CCW from r direction to F Sign of Torque If the torque causes a counterclockwise angular acceleration, it is positive; if it causes a clockwise angular acceleration, it is negative. CCW: τ > 0 CW: τ < 0 Lecture 22 7/26 Lecture 22 8/26 direction.)

Question Five different forces are applied to the same rod, which pivots around the black dot. Which force produces the smallest torque about the pivot?

3 Question Five different forces are applied to the same rod, which pivots around the black dot. Which force produces the smallest torque about the pivot? 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 = Mgx grav cm Lecture 22 9/26 Lecture 22 10/26 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? Example: Determine the net torque: N 4 m 2 m τ = = = 2 grav Mgxcm (500 kg)(9.80 m/s )( 0.80m) 3920 Nm 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 Lecture 22 11/26 Rotation would be CCW Lecture 22 12/26

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: Torque & Angular Acceleration Once

4 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: Torque & Angular Acceleration Once again, we have analogies between linear and angular motion: Or equivalently, F = m a τ = I α Lecture 22 13/26 Lecture 22 14/26 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? τ Stationary Bike net = F r = Iα τ I net α = = s Fr s MR 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 22 15/26 Lecture 22 16/26

Example: Uniform Rod Pivoted at End A uniform thin rod of length L and mass M is pivoted at one end.

Find angular acceleration α of rod immediately after its release.

object with forces acting on it, but that is not moving (or rotating), is said to be in equilibrium.

5 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 = Iα ext τ = Mg grav 2 1 ( L ) 1 ( 2 ) α = grav = 1 2 = 3 I = τ Mg L 3g I ML 2L ML Lecture 22 17/26 Static Equilibrium An object with forces acting on it, but that is not moving (or rotating), is said to be in equilibrium. Lecture 22 18/26 Conditions for Equilibrium The first condition for equilibrium is that the forces along each coordinate axis add to zero. Conditions for Equilibrium The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary. Lecture 22 19/26 Lecture 22 20/26

Often choose axis through which unknown force acts.

6 Zero Torque & Static Equilibrium Static equilibrium occurs when an object is at rest, neither rotating nor translating. 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. Lecture 22 21/26 Axis Axis Lecture 22 22/26 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. ConcepTest Balancing Rod 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? 1m 1) 1/4 kg 2) 1/2 kg 3) 1 kg 4) 2 kg 5) 4 kg 1kg Lecture 22 23/26 Lecture 22

7 Example 2: What distance d 2 should girl have from axis for equilb m? What will be normal force N? (Neglect see-saw mass.) d 2 2 m N y 500 N 800 N 1. Draw all applicable forces and lever arms τ RHS = (800 N)(2 m) 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 22 25/26 Equilibrium 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 22 26/26 Torque Condition: L τ = mg cos 50 PL sin 50 = 0 2 P sin 50 = (25N ) cos 50 P = 21N Force Conditions: F f = P = 21N F x y = 0 = f P = 0 = n mg n = mg = 50N f µ s nso need µ s (f/n) or µ s 0.42 Lecture 22 27/26 End of Lecture 22 For Friday, read Walker, 11.3; Homework Assignment 11a is due at 11:00 PM on Sunday, Nov. 1. Lecture 22 28/26

Physics 111. Lecture 23 (Walker: 10.6, 11.1) Conservation of Energy in Rotation Torque March 30, Kinetic Energy of Rolling Object

Physics 111. Lecture 23 (Walker: 10.6, 11.1) Conservation of Energy in Rotation Torque March 30, Kinetic Energy of Rolling Object Physics 111 Lecture 3 (Walker: 10.6, 11.1) Conservation of Energy in Rotation Torque March 30, 009 Lecture 3 1/4 Kinetic Energy of Rolling Object Total kinetic energy of a rolling object is the sum of