A string is wrapped around a uniform disk of mass M = 1.8 kg and radius R = 0.11

Question

A string is wrapped around a uniform disk of mass M = 1.8 kg and radius R = 0.11 m (see figure below). Attached to the disk are four low-mass rods of radius b = 0.13 m, each with a small mass m = 0.6 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 = 22 N for a time of 0.2 s.

a/ Now what is the angular speed of the apparatus? (rad/s)

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

Explanation / Answer

Moment of inertia of assembly:

I = ½MR² + 4mb²

I = ½(1.8kg)(0.11m)² + 4(0.6kg)(0.13m)² = 0.05145 kg·m²

torque ? = I*?

and also ? = F*R = 22N * 0.11m = 2.42 N·m

so

2.42 N·m = 0.05145kg·m² * ?

? = 47.04 rad/s²

(a) ? = ?*t = 47.04rad/s² * 0.2s = 9.41 rad/s ?

(b) ? = ½?t² = ½ * 47.04rad/s² * (0.2s)² = 0.9408 radians ?

which is 0.1497 rev and 53.9º ?

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