A string is wrapped around a uniform disk of mass M = 1.7 kg and radius R = 0.11

Question

A string is wrapped around a uniform disk of mass M = 1.7 kg and radius R = 0.11 m (see figure below). Attached to the disk are four low-mass rods of radius b = 0.17 m, each with a small mass m = 0.5 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F= 16 N for a time of 0.2 s. Now what is the angular speed of the apparatus?

Also calculate numerically the angle through which the apparatus turns, in radians and degrees.

Also calculate numerically the angle through which the apparatus turns, in radians and degrees.

Explanation / Answer

torque applied by Force F is T = R*F*sin(90) = 0.11*16*1 = 1.76 N-m

Let I be the moment of inertia of the system

I = 0.5*M*R^2 + (4*m*b^2) = (0.5*1.7*0.11^2)+(4*0.5*0.17^2) = 0.068085 kg-m^2

using T = I*alpha

1.76 = 0.068085*alpha

alpha = 1.76/0.068085 = 25.85 rad/s^2

but alpha (Wf-Wi)/t = 25.85

initial angular velocity is Wi = 0 rad/s

Final angular velocity is Wf = ?

time taken is t = 0.2 sec

then alpha = 25.85 = (Wf-0)/0.2

Wf = 5.17 rad/s

then angular displacemnt is theta = 0.5*alpha*t^2 = 0.5*25.85*0.2^2 = 0.517 rad

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