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Sagot :
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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