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A ball starts at the top of a hill that is angled 25 degrees above the horizontal and has a length of 40m. If the ball has a mass of 5.0 kg how long does it take the ball to reach the bottom of the hill? What velocity does the ball have when it reaches the bottom of the hill?

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Answer:

Taking torque about the point of contact:

M g  R sin theta = 7 / 5  M R^2  * alpha

(used the parallel axis theorem to get torque about point of contact)

One can also take torque about center of mass, but then the force of friction must be considered

alpha * R = 5/7 g sin 25 = a = .302 g  = 2.96 m/s^2  linear acceleratkon

S = 1/2 a t^2

t = (2 S / a)^1/2 = (80 / 2.96)^1/2 = 5.20 s

V = a t = 2.96 ms^2 * 5.2 s = 15.4 m/s

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