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Standard XII

Physics

Applying NLM along 3 Axis

A rigid body ...

Question

A rigid body of radius 'R' and radius of gyration 'K' is rolling down along a rough inclined plane of inclination $θ$ from height 'h'. It's linear acceleration is _________.

\frac{g sin θ}{\frac{k^{2}}{R^{2}} - 1}

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\frac{g sin θ}{\frac{R^{2}}{K^{2}} - 1}

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\frac{g sin θ}{1 + \frac{R^{2}}{K^{2}}}

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\frac{g sin θ}{1 + \frac{K^{2}}{R^{2}}}

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Solution

The correct option is D $\frac{g sin θ}{1 + \frac{K^{2}}{R^{2}}}$
Equations are:
$m g s i n θ - f = m a . . . . (1) f R = m K^{2} α . . . . (2) a = R α . . . . . (3)$

Solving these equations we get:
$a = \frac{g s i n θ}{1 + \frac{K^{2}}{R^{2}}}$

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A rigid body of radius 'R' and radius of gyration 'K' is rolling down along a rough inclined plane of inclination θ from height 'h'. It's linear acceleration is _________.

The correct option is D gsinθ1+K2R2Equations are:mgsinθ−f=ma....(1)fR=mK2α....(2)a=Rα.....(3)Solving these equations we get:a=gsinθ1+K2R2

A rigid body of radius 'R' and radius of gyration 'K' is rolling down along a rough inclined plane of inclination $θ$ from height 'h'. It's linear acceleration is _________.

The correct option is D $\frac{g sin θ}{1 + \frac{K^{2}}{R^{2}}}$
Equations are:
$m g s i n θ - f = m a . . . . (1) f R = m K^{2} α . . . . (2) a = R α . . . . . (3)$

Solving these equations we get:
$a = \frac{g s i n θ}{1 + \frac{K^{2}}{R^{2}}}$