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JEE Main Syllabus 2025 PDF (Out) for Physics, Chemistry, Maths

Rolling Without Slipping On An Inclined Plane - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 12 Questions around this concept.

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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)

Concepts Covered - 1

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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Rolling without slipping on an Inclined Plane

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Rolling without slipping on an Inclined Plane

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