M, a solid cylinder (M=1.47 kg, R=0.137 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.47 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass, i.e., F = 6.965 N.Calculate the angular acceleration of the cylinder. If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder. How far does m travel downward between 0.690 s and 0.890 s after the motion begins? The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.600 m in a time of 0.570 s. Find Icm of the new cylinder.

Explanation / Answer

part a )

torque = F*R = I*alpha

I = mR^2 = 0.0138 kg.m^2

alpha = F*R/I = 69.145 rad/s^2

Part B )

F = mg - ma

t = F*R

I*alpha = F*R

I*(a/R) = F*R

a = mg/(I/R^2 + m )

a = 6.965/(0.0138/0.137^2 + 1.47)

a = 3.16 m/s^2

alpha = a/r = 23.05 rad/s^2

part c )

s = 1/2 * a*t^2

s = 1/2* 3.16 ( 0.890^2 - 0.690^2)

s = 0.49928 = 0.5 m

part d )

s = 1/2 * a *t^2

0.600 = 1/2 * a * .570^2

a = 3.69 m/s^2

a = mg /(I/R^2 + m)

3.69 = 6.965/(I/0.137^2 + 1.47)

(I/0.137^2 + 1.47) = 9.965/3.69

I = 0.023 kg .m^2

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