SATHEE: Rotation 3 Question 8

Rotation 3 Question 8

14. The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

(a) $- k x$

(b) $- 2 k x$

(c) $- \frac{2 k x}{3}$

(d) $- \frac{4 k x}{3}$

(2008, 4M)

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Answer:

Correct Answer: 14. (d)

Solution:

$\begin{aligned} a & = R α \\ ∴ \frac{2 k x - f}{M} & = R \frac{f R}{\frac{1}{2} M R^{2}} \end{aligned}$

Solving this equation, we get

$\begin{aligned} f & = \frac{2 k x}{3} \\ ∴ | F_{net} | & = 2 k x - f = 2 k x - \frac{2 k x}{3} = \frac{4 k x}{3} \end{aligned}$

This is opposite to displacement.

$∴ F_{net} = - \frac{4 k x}{3}$

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Rotation 3 Question 8

14. The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is

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14. The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is