PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 12
Last Lecture Newton s Law of gravitation F grav = GMm R 2 G = 6.67"10 #11 Nm 2 /kg 2 Kepler s Laws of Planetary motion 1. Ellipses with sun at focus 2. Sweep out equal areas in equal times 3. R 3 T 2 = GM 4" 2 = Constant
Gravitational Potential Energy PE = mgh valid only near Earth s surface For arbitrary altitude PE =!G Mm r Zero reference level is at r=!
Example 7.18 You wish to hurl a projectile from the surface of the Earth (R e = 6.38x10 6 m) to an altitude of 20x10 6 m above the surface of the Earth. Ignore rotation of the Earth and air resistance. a) What initial velocity is required? a) 9,736 m/s b) What velocity would be required in order for the projectile to reach infinitely high? I.e., what is the escape velocity? b) 11,181 m/s c) (skip) How does the escape velocity compare to the velocity required for a low earth orbit? c) 7,906 m/s
Chapter 8 Rotational Equilibrium and Rotational Dynamics
Wrench Demo
Torque Torque, ", is tendency of a force to rotate object about some axis! = Fd F is the force d is the lever arm (or moment arm) Units are Newton-meters Door Demo
Torque is vector quantity Direction determined by axis of twist Perpendicular to both r and F Clockwise torques point into paper. Defined as negative Counter-clockwise torques point out of paper. Defined as positive r r F F - +
Non-perpendicular forces! = Fr sin"! is the angle between F and r
Torque and Equilibrium Forces sum to zero (no linear motion)!f x = 0 and!f y = 0 Torques sum to zero (no rotation)!" = 0
Axis of Rotation Torques require point of reference Point can be anywhere Use same point for all torques Pick the point to make problem easiest (eliminate unwanted Forces from equation)
Example 8.1 Given M = 120 kg. Neglect the mass of the beam. a) Find the tension in the cable b) What is the force between the beam and the wall a) T=824 N b) f=353 N
Another Example Given: W=50 N, L=0.35 m, x=0.03 m Find the tension in the muscle W x L F = 583 N
Center of Gravity Gravitational force acts on all points of an extended object However, one can treat gravity as if it acts at one point: the center-of-gravity. $ #m " = # i x i (m i g)x i = M tot g & % M tot ' ) = M gx tot ( Center of gravity: X = "m i x i M tot
Example 8.2 Given: x = 1.5 m, L = 5.0 m, w beam = 300 N, w man = 600 N Find: T Fig 8.12, p.228 Slide 17 T = 413 N x L
Example 8.3 Consider the 400-kg beam shown below. Find T R T R = 1 121 N
Example 8.4a T left T right Given: W beam =300 W box =200 Find: T left W beam A B C D 8 m 2 m W box What point should I use for torque origin? A B C D
Example 8.4b T left T right Given: T left =300 T right =500 Find: W beam W beam A B C D 8 m 2 m W box What point should I use for torque origin? A B C D
Example 8.4c T left T right Given: T left =250 T right =400 Find: W box W beam A B C D 8 m 2 m W box What point should I use for torque origin? A B C D
Example 8.4d T left T right Given: W beam =300 W box =200 Find: T right W beam A B C D 8 m 2 m W box What point should I use for torque origin? A B C D
Example 8.4e T left T right Given: T left =250 W beam =250 Find: W box W beam A B C D 8 m 2 m W box What point should I use for torque origin? A B C D
Torque and Angular Acceleration Analogous to relation between F and a F = ma,! = I" F m Moment of Inertia R " = FR = (ma t )R = m(#r)r " = mr 2 #
Moment of Inertia Moment of inertia, I: rotational analog to mass I = m i r i 2! i r defined relative to rotation axis SI units are kg m 2
Baton Demo Moment-of-Inertia Demo
More About Moment of Inertia Depends on mass and its distribution. If mass is distributed further from axis of rotation, moment of inertia will be larger.
Moment of Inertia of a Uniform Ring Divide ring into segments The radius of each segment is R I =!m i r i 2 = MR 2
Example 8.6 What is the moment of inertia of the following point masses arranged in a square? a) about the x-axis? b) about the y-axis? c) about the z-axis? a) 0.72 kg#m 2 b) 1.08 kg#m 2 c) 1.8 kg#m 2
Other Moments of Inertia
Other Moments of Inertia cylindrical shell : I = MR 2 solid cylinder : I = 1 2 MR2 rod about center : I = 1 12 ML2 rod about end : I = 1 3 ML2 bicycle rim filled can of coke baton baseball bat spherical shell : I = 2 3 MR2 basketball solidsphere : I = 2 5 MR2 boulder
Example 8.7 Treat the spindle as a solid cylinder. a) What is the moment of Inertia of the spindle? (M=5.0 kg, R=0.6 m) b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? M d) What is the mass of the bucket? a) 0.9 kg#m 2 b) 6.67 rad/s 2 c) 4 m/s 2 d) 1.72 kg
Example 8.9 A 600-kg solid cylinder of radius 0.6 m which can rotate freely about its axis is accelerated by hanging a 240 kg mass from the end by a string which is wrapped about the cylinder. a) Find the linear acceleration of the mass. 4.36 m/s 2 b) What is the speed of the mass after it has dropped 2.5 m? 4.67 m/s