Review from Last Lecture

Review from Last Lecture • Newton’s Laws • Inertial frames exist • F = ma • F12 = -F21 • The Normal Force • Perpendicular to surface • Large enough to keep objects from interpenetrating • Friction • Parallel to surface, resisting motion • fs£msn: Resists any motion up to the amount specified • fk = mkn: Works against motion when object is moving

Springs • Springs exert a force depending on how far they are stretched • The magnitude of the force is proportional to how far the spring is stretched or compressed from equilibrium • The direction of the force is opposite to the direction of stretching or compression • F = -k Dx = -k (x-x0) where x0 is the equilibrium position • Often choose x0 = 0 so F = -k x

Ropes and Pulleys T T T T • Inside a rope, net force is zero at every point • At ends, rope exerts a force T due to tension • Can change direction of tension with a pulley T T

Atwood’s Machine • What are the acceleration and the tension? • Each mass subject to gravity and tension • Define y-axis sign so that a is identical for both masses (a is identical because of rope) • Up positive for m1 • Down positive for m2 • Thus T-m1g = m1aÞ T = m1g + m1a m2g-T = m2aÞ T = m2g – m2a m1g + m1a = m2g – m2a (m1+ m2) a = (m2– m1) g a = (m1– m1) g / (m1+ m2) T = m1(g + a) = 2m1m2 g / (m1+ m2)

Problem Solving Examples A block of mass m1 on a rough, horizontal surface ks connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown. A force of magnitude F at angle q is applied to the block. The coefficient of friction between the block and the surface is mk. Determine the magnitude of acceleration of the two objects.

Problem Solving Examples • Draw the objects • Draw all the forces acting on the objects n T fk T m1g m2g

Problem Solving Examples • Draw the object • Draw all the forces acting on the object • Choose a coordinate system • Find the components of all the forces in that coordinate system n F sinq T F cosq fk T m1g m2g

Problem Solving Examples • Draw the object • Draw all the forces acting on the object • Choose a coordinate system • Find the components of all the forces in that coordinate system • Solve SF = ma for each coordinate n F sinq T F cosq fk T m1g m2g

Problem Solving Examples • Solve SF = ma for each coordinate • SFx = Fcosq – T – fk = Fcosq – T – mkn = m1a SFy = n + Fsinq - m1g= 0 Þn = m1g - Fsinq • SFx = 0 SFy = T – m2g = m2a ÞT = m2(g +a) m1a = Fcosq – m2(g+a) – mk(m1g-Fsinq ) a(m1+m2) = Fcosq–m2g–mk(m1g-Fsinq) n F sinq T F cosq fk T m1g m2g

Uniform Circular Motion (again!) • In Lecture 3 we learned that for uniform circular motion • Now we study the force that keeps an object moving in UCM • Since by Newton’s 2nd we have F=ma, clearly

Uniform Circular Motion • What if we cut the sting? • The ball should move off with constant velocity • This means the ball will continue along the tangent to the circle, not move away from the center radially (with constant angular velocity)

Uniform Circular Motion • The Conical Pendulum • As the ball revolves faster, the angle increases • What’s the speed for a given angle?

Uniform Circular Motion

UCM Example: Galactic Rotations

UCM Example: Galactic Rotations • Stars orbit their galaxies in circles at various speeds • Speeds measured by Doppler shifts in the light

UCM Example: Galactic Rotations • Stars orbit their galaxies in circles at various speeds • Speeds measured by Doppler shifts in the light • Gravity predicts:

UCM Example: Galactic Rotations • Stars orbit their galaxies in circles at various speeds • Speeds measured by Doppler shifts in the light • Gravity predicts: v=(GM/R)0.5 • But outer objects move too fast. Need dark matter! Measured v vs R Expected v vs R

(Non-) Uniform Circular Motion • What about swinging the ball about vertically? • The speed is no longer constant! • What is tension? • Use radial coordinate system

(Non-) Uniform Circular Motion • What’s the slowest the ball can go at top, and stay on circle?

Review from Last Lecture