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Q.

A solid cylinder of mass 3 kg is placed on a rough inclined plane of inclination 30^o. If $g = 10 m s^{- 2}$ , then the minimum frictional force required for it to roll without slipping down the plane is

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a

15N

b

18N

c

5N

d

2N

answer is B.

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Detailed Solution

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$\begin{array}{l} τ = f R = I α or f R = \frac{1}{2} m R^{2} \times \frac{a}{R} \\ ⇒ f = \frac{1}{2} m a . . . . (i) \\ Now, m g \sin θ - f = m a \end{array}$

$\begin{array}{l} ⇒ m g \sin θ - \frac{1}{2} m a = m a \\ ⇒ \frac{3}{2} m a = m g \sin θ or \frac{1}{2} m a = \frac{m g \sin θ}{3} \\ From equation (i) f = \frac{m g \sin θ}{3} \\ Substituting values, f = \frac{3 \times 10 \times 0.5}{3} N = 5 N \end{array}$

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