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A bowling ball (mass 7.00 kg and radius 17.0 cm) is released so fast that it skids without rotating down the alley (at least for a while). Assume the ball skids to your right and the coefficient of sliding friction between the ball and the lane surface is 0.400. (a) What is the direction of the torque exerted by the friction on the ball about the center of mass of the ball? [ clockwise] (b) determine the magnitude of this torque (again about the ball’s center of mass) [

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jolis1796

Answer:

a) Clockwise

b) τ = 4.67 N-m

Explanation:

Given

m = 7 Kg

R = 17 cm = 0.17 m

μ = 0.4

We use the formula

τ = R*F*Sin ∅ = R*(μ*m*g)*Sin ∅

⇒  τ = (0.17 m)*(0.4*7 Kg*9.81 m/s²)*Sin 270° = - 4.67 N-m

(Clockwise)

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