Question

M, a solid cylinder (M=1.39 kg, R=0.115 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 0.830 kg mass, i.e., F = 8.142 N.

1.) Calculate the angular acceleration of the cylinder.

2.) If instead of the force F an actual mass m = 0.830 kg is hung from the string, find the angular acceleration of the cylinder.

3.) How far does m travel downward between 0.730 s and 0.930 s after the motion begins?

4. 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 0.566 m in a time of 0.510 s. Find I_cm of the new cylinder.

Answer 1

1)

net torque = I*alpha

F*r = (1/2)*M*r^2*alpha

alpha = 2*F/(r*M)

alpha = 2*8.142/(0.115*1.39)

alpha = 102 rad/s^2 <<<-------------answer

=---------------------

(2)

for the cylinder

net torque = I*alpha

T*r = (1/2)*M*r^2*a/r

T = (1/2)*M*a

for mass

mg - T = m*a

m*g - (1/2)*M*a = m*a

acceleration a = 2*m*g/(M + 2m)

a = 2*0.83*9.8/(1.39 + (2*0.83))

a = 5.33 m/s^2

angular accelration alpha = a/r = 5.33/0.115 = 46.35 rad/s^2 <<<-------------answer

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3)

for the mass

distance = dy = (1/2)*a*(t2^2-t1^2)

dy = (1/2)*5.33*(0.93^2-0.73^2) = 0.885 m   <<<-------------answer

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4)

acceleration a = 2*y/t^2 = 2*0.566/0.51^2 = 4.35 m/s^2

for the cylinder

Fnet = I*alpha

T*r = I*a/r

(mg - ma)*r = I*a/r

I = m*(g-a)*r^2/a

I = 0.83*(9.81-4.35)*0.115^2/4.35

I = 0.0138 kg m^2 <<<-------------answer