Rolling Without Slipping On An Inclined Plane MCQ

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A solid sphere, a hollow sphere and a ring are released from the top of an inclined plane ( (frictionless ) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)

Solid sphere

hollow sphere

ring

all same

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Rolling without slipping on an Inclined Plane

When a body of mass m and radius R rolls down an inclined plane having an angle of inclination ( $\theta$ ) and at height ‘h’

By conservation of mechanical energy

$mgh=\frac{1}{2}mV^2(1+\frac{K^2}{R^2})$

Where V=Velocity at the lowest point

And, $V=\sqrt{\frac{2gh}{1+\frac{K^2}{R^2}}}$

Similarly using $V^2=u^2+2as$

$\text{Acceleration}=a= \frac{g\sin \Theta }{1+\frac{K^{2}}{R^{2}}}$

And angular acceleration = $a =R\alpha$

And we know that $\tau = I\alpha$

And torque due to friction force = $\tau _f=fR=I\alpha =mK^2 (Ra)$

So, $f= \frac{mg\sin \Theta }{1+\frac{R^{2}}{K^{2}}}$

As $f= \mu N= \mu mgcos\theta$

So Condition for pure rolling on an inclined plane

$\mu _{s}\geq \frac{\tan \Theta }{1+\frac{R^{2}}{K^{2}}}$

Where $\mu _{s}$ = limiting coefficient of friction

And let t= time taken by the body to reach the lowest point

So using $V = u+at$

We get, $\mathbf{t=\frac{1}{sin\theta }\sqrt{\frac{2h}{g}\left (1+\left [\frac{K^2}{R^2} \right ] \right )}}$

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