M, a solid cylinder (M=1.47 kg, R=0.133 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.47 kg, R=0.133 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. The answer is 7.13 x 10^1

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. The answer is 3.62 rad/s^2

How far does m travel downward between 0.550 s and 0.750 s after the motion begins? The answer is 6.27 x10^-1

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 0.577 m in a time of 0.570 s. Find Icm of the new cylinder. I cannot figure this one out all of the following answers are incorrect: .333 kgm^2, 8.64 x 10^-3, .0347, .0444

Explanation / Answer

mass moves a distance of 0.577 m in t = 0.570 sec

now

acceleration = 2*s/t^2

a = 2*0.577/0.57^2 = 3.55 m/sec^2

F = mg - ma

F = 6.965 - 0.71*3.55 = 4.4445 N

Torque balance

F*r = I*alpha

alpha = a/r

I = F*r/alpha

I = F*r^2/a

I = 4.4445*0.133^2/3.55 = 0.0221 kg-m^2

Comment below If you have any doubt.

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