A spherical bowling ball with mass m = 4 kg and radius R = 0.108 m is thrown dow

Question

A spherical bowling ball with mass m = 4 kg and radius R = 0.108 m is thrown down the lane with an initial speed of v = 8.8 m/s. The coefficient of kinetic friction between the sliding ball and the ground is = 0.32. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

1.) How long does it take the bowling ball to begin rolling without slipping?

2.) How far does the bowling ball slide before it begins to roll without slipping?

3.) What is the magnitude of the final velocity?


can someone please explain the steps on how to solve this please?

Explanation / Answer

The equations of motion for translation and rotation are: m dv/dt = -µ m g I dw/dt = µ m g R The first is solved by v(t) = v0 - µ g t As for the second, using I = 2/5 m R^2 for a solid uniform sphere (ignoring the holes in the bowling ball), this is solved by w(t) = w0 + 5 µ g t / (2R) Therefore 1. angular acceleration is 5 µ g /(2 R) 2. linear acceleration is -µ g 3. When is v(t) = w(t) R (pure rolling)? v0 - µ g t = w0 R + 5 µ g t / 2 (7 µ g /2) t = v0 + w0 R t =( 2 v0 + 2 w0 R)/( 7 µ g ) Assuming w0 = 0 (no spin is given in the throw) , this is t = 2 v0 / (7 µ g ) 4. s(t) = v(0) t + 1/2 a t^2 = 2 v0^2 / (7 µ g ) - 1/2 * (µ g) ( 4 v0^2 / (7µ g )^2 = 2/7 (v0^2 / (µ g) ) - 2/49 (v0^2 / (µ g) ) = 12 v0^2/( 49 µ g) numerical given : 1. angular acceleration = 5*0.32*9.81 m/s^2 / ( 2 * 0.108 m) = 72.6 radians/second^2 1. linear acceleration = - 0.32 * 9.81 m/s^2 = -3.13 m/s^2 2. t = 2*8.8 m/s / ( 7 * 0.32 * 9.81 m/s^2) = 0.8 s 3. s = 12 * (8.8 m/s)^2 / (49 * 0.32 * 9.81 m/s^2) = 6.04 m

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