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PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 12 Last Lecture GMm Fgrav = R2 G = 6.67 "10#11 Nm 2 /kg 2 • Newton’s Law of gravitation • Kepler’s Laws of Planetary motion ! 1. Ellipses with sun at focus 2. Sweep out equal areas in equal times 3 R GM 3. = = Constant 2 2 T 4" ! Gravitational Potential Energy • PE = mgh valid only near Earth’s surface • For arbitrary altitude Mm PE = !G r • Zero reference level is at r=! Example 7.18 You wish to hurl a projectile from the surface of the Earth (Re= 6.38x106 m) to an altitude of 20x106 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) !Fx = 0 and !Fy = 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 x ' i i & ) = M tot gX " = # (mi g)x i = M tot g& ) M tot % ( • Center of gravity: ! ! mx " X= i M tot i Example 8.2 Given: x = 1.5 m, L = 5.0 m, wbeam = 300 N, wman = 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 TR TR = 1 121 N Example 8.4a Tleft Wbeam B A Given: Wbeam=300 Wbox=200 Find: Tleft Tright D C 8m 2m Wbox What point should I use for torque origin? A B C D Example 8.4b Tleft Wbeam B A Given: Tleft=300 Tright=500 Find: Wbeam Tright D C 8m 2m Wbox What point should I use for torque origin? A B C D Example 8.4c Tleft Wbeam B A Given: Tleft=250 Tright=400 Find: Wbox Tright D C 8m 2m Wbox What point should I use for torque origin? A B C D Example 8.4d Tleft Wbeam B A Given: Wbeam=300 Wbox=200 Find: Tright Tright D C 8m 2m Wbox What point should I use for torque origin? A B C D Example 8.4e Tleft Wbeam B A Given: Tleft=250 Wbeam=250 Find: Wbox Tright D C 8m 2m Wbox 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" Moment of Inertia F m R " = FR = (mat )R = m(#R)R " = mR 2# ! Moment of Inertia • Moment of inertia, I: rotational analog to mass I = ! mi ri2 i • r defined relative to rotation axis • SI units are kg m2 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 = !mi ri2 = 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#m2 b) 1.08 kg#m2 c) 1.8 kg#m2 Other Moments of Inertia Other Moments of Inertia cylindrical shell : I = MR 2 1 solid cylinder : I = MR 2 2 1 rod about center : I = ML2 12 1 rod about end : I = ML2 3 2 2 spherical shell : I = MR 3 2 solidsphere : I = MR 2 5 bicycle rim filled can of coke baton baseball bat basketball 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#m2 b) 6.67 rad/s2 c) 4 m/s2 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/s2 b) What is the speed of the mass after it has dropped 2.5 m? 4.67 m/s