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Figure shows a spool with thread wound on it placed on a smooth plane inclined at angle from horizontal. The spool has mass m, edge radius R, and is wound up to a radius r. its moment of inertia about its own axis is I. The free end of the thread is attached as shown in the figure. So that the thread is parallel to the inclined plane. T is the tension in the thread. Which of the following is correct?

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The linear acceleration of the spool axis down the slope is $\frac{mgsin θ - T}{m}$ .

Angular acceleration is $\frac{Tr}{2 I}$ .

The linear acceleration of the spool axis down the plane is $\frac{{Tr}^{2}}{I}$ .

The acceleration of the spool axis down the slope is $\frac{gsin θ}{1 + \frac{I}{{mr}^{2}}}$ .

answer is A, C, D.

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Detailed Solution

$Linear acceleration =a = \frac{mgsin θ - T}{m}$

$Angular acceleration = \frac{Torque (Tr)}{Moment of inertia (I)} = \frac{Tr}{I}$

$Linear acceleration=a=rα = r [\frac{Tr}{I}] = \frac{{Tr}^{2}}{I}$

Since the two linear accelerations must match,

$∴ \frac{{Tr}^{2}}{I} = \frac{mgsin θ - T}{m} or \frac{{Tmr}^{2}}{I} + T = mgsin θ$

$or T [1 + \frac{{mr}^{2}}{I}] = mgsin θ or T = \frac{mgsin θ}{1 + \frac{{mr}^{2}}{I}}$

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