Math Review: Circular Motion 8.01

Transcription

1 Math Review: Circular Motion 8.01

2 Position and Displacement r ( t) : position vector of an object moving in a circular orbit of radius R Δr ( t) : change in position between time t and time t+δt Position vector is changing in direction not in magnitude. The magnitude of the displacement is the length of the chord of the circle: Δ r = 2Rsin( Δθ / 2)

3 Direction of Velocity Sequence of chord Δr directions approach direction of velocity as Δt approaches zero. The direction of velocity is perpendicular to the direction of the position and tangent to the circular orbit. Direction of velocity is constantly changing.

4 Small Angle Approximation When the angle is small: Power series expansion sinφ φ, cosφ 1 sinφ = φ φ 3 3! + φ5 5!... cosφ = 1 φ 2 2! + φ 4 4!... Using the small angle approximation with φ = Δθ / 2, the magnitude of the displacement is Δ r = 2Rsin( Δθ / 2) R Δθ

5 Speed and Angular Speed The speed of the object undergoing circular motion is proportional to the rate of change of the angle with time: Δr R Δθ Δθ dθ v v = lim = lim = R lim = R = Rω Δt 0 Δt Δt 0 Δt Δt 0 Δt dt Angular speed: ω = dθ dt (units: rad s -1 )

6 Circular Motion: Constant Speed, Period, and Frequency In one period the object travels a distance equal to the circumference: s = 2π R = vt Period: the amount of time to complete one circular orbit of radius R 2π R 2π R 2π T = = = v Rω ω Frequency is the inverse of the period: f 1 ω = = T 2π (units: s 1 or Hz)

7 Checkpoint Problem: Ball and spring One end of a spring is attached to the central axis of a motor. The axis of the motor is in the vertical direction. A small ball of mass m 2 is then attached to the other end of the spring. The motor rotates at a constant frequency f. Neglect the gravitational force exerted on the ball. Assume that the ball and spring rotate in a horizontal plane. The spring constant is k. Let r 0 denote the unstretched length of the spring. (i) How long does it take the ball to complete one rotation? (ii) What is the angular frequency of the ball in radians per sec?

8 Acceleration and Circular Motion When an object moves in a circular orbit, the direction of the velocity changes and the speed may change as well. For circular motion, the acceleration will always have a radial component (a r ) due to the change in direction of velocity The acceleration may have a tangential component if the speed changes (a t ). When a t =0, the speed of the object remains constant

9 Direction of Radial Acceleration: Uniform Circular Motion Δv Sequence of chord directions approaches direction of radial acceleration as Δt approaches zero Perpendicular to the velocity vector Points radially inward

10 Checkpoint Problem: Magnitude of Change in Velocity for Circular Motion Find the change in the magnitude of the velocity in terms of v and Δθ. Use the small angle approximation as necessary.

11 Magnitude of Change in Velocity: Circular Motion Solution Change in velocity: Δ v = v( t + Δt) v( t) Magnitude of change in velocity: Δ v = 2vsin ( Δθ / 2) Using small angle approximation Δv vδθ

12 Radial Acceleration: Constant Speed Circular Motion Any object traveling in a circular orbit with a constant speed is always accelerating towards the center. Direction of velocity is constantly changing. Radial component of a r (minus sign indicates direction of acceleration points towards center) a r = lim Δt 0 Δv Δt = lim vδθ Δθ dθ v2 = v lim = v = vω = Δt 0 Δt Δt 0 Δt dt R a r = v2 R = ω 2 R

13 Tangential Acceleration The tangential acceleration is the rate of change of the magnitude of the velocity 2 Δv Δω dω d θ at = lim = R lim = R = R = Rα Δt 0 t 0 2 Δt Δ Δt dt dt Angular acceleration: rate of change of angular velocity with time α = dω dt = d 2 θ dt 2 (units: rad s -2 )

14 Alternative forms of Magnitude of Radial Acceleration Parameters: speed v, angular speed ω, angular frequency f, period T v ar = = R = R f = R 4π T ω (2 π ) 2 R

15 Checkpoint Problem: Spinning Earth The earth is spinning about its axis with a period of 23 hours 56 min and 4 sec. The equatorial radius of the earth is m. The latitude of MIT in Cambridge, Mass is 42 o 22. a) Find the velocity of a person at MIT as they undergo circular motion about the earth s axis of rotation. b) Find the person s centripetal acceleration.

16 Summary: Kinematics of Arc length Circular Motion s = Rθ Tangential velocity Tangential acceleration Centripetal v = ds dt = R dθ dt = Rω 2 dv d θ at = = R = Rα 2 dt dt 2 v ar = vω = = Rω R 2

17 Cylindrical Coordinate System Coordinates Unit vectors (r,θ, z) ( rˆ, θˆ, zˆ )

18 Circular Motion: Vector Description Use plane polar coordinates Position Angular Speed Velocity Angular Acceleration Acceleration r(t) = R ˆr(t) ω dθ / dt v = v θ ˆθ v(t) = R(dθ / dt) ˆθ = Rω ˆθ α dω / dt = d 2 θ / dt 2 a = a r ˆr + a θ ˆθ a r = rω 2 = (v 2 / r), a θ = rα

19 Newton s Second Law: Equations of Motion for Circular Motion

20 Concept Question: Circular Motion and Force A pendulum bob swings down and is moving fast at the lowest point in its swing. T is the tension in the string, W is the gravitational force exerted on the pendulum bob. Which free-body diagram below best represents the forces exerted on the pendulum bob at the lowest point? The lengths of the arrows represent the relative magnitude of the forces.

21 Checkpoint Problem: Ball and spring One end of a spring is attached to the central axis of a motor. The axis of the motor is in the vertical direction. A small ball of mass m 2 is then attached to the other end of the spring. The motor rotates at a constant frequency f. Neglect the gravitational force exerted on the ball. Assume that the ball and spring rotate in a horizontal plane. The spring constant is k. Let r 0 denote the unstretched length of the spring. (i) How long does it take the ball to complete one rotation? (ii) What is the angular frequency of the ball in radians per sec? (iii) What is the radius of the circular motion of the ball?