A rigid disk is free to rotate about a fixed axis through its center

PHYSICS · Multi-part · CBSE · Physics · System of Particles and Rotational Motion · System of Particles and Rotational Motion

Problem

A rigid disk is free to rotate about a fixed axis through its center. A mechanism applies a torque that varies with angular position

\theta

(in radians) according to the table below.

Interval in $\theta$	Constant torque within interval ( $\text{N}\cdot\text{m}$ )
$0 \le \theta \le \frac{\pi}{3}$	$+10$
$\frac{\pi}{3} < \theta \le \frac{\pi}{2}$	$+6$
$\frac{\pi}{2} < \theta \le \pi$	$-4$

Positive torque corresponds to the counter-clockwise direction. The disk starts from

\theta = 0

. Neglect all frictional effects. (a) Determine the net work done on the disk as it rotates from

\theta = 0

\theta = \pi

. (b) At what angular position within the given range (if any) is the cumulative work on the disk equal to zero for the first time? Show how you arrive at your answer. (c) Suppose the disk has a moment of inertia of

0.40\,\text{kg}\cdot\text{m}^{2}

and starts from rest. Calculate its angular speed after it reaches

\theta = \pi

, using your result from part (a). (d) Interpret physically why the portion of the motion with negative torque affects the final angular speed the way it does, citing the sign of the work in that interval.

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