A round uniform body of radius     mass  M  and moment of inertia  I   rolls  down ( without slipping ) an inclined plane making an angle \Theta   with the horizontal. Then its acceleration is

Option 1)

\frac{g\sin \Theta }{1-MR^{2}/I}

Option 2)

\frac{g\sin \Theta }{1+I/MR^{2}}

Option 3)

\frac{g\sin \Theta }{1+MR^{2}/I}

Option 4)

\frac{g\sin \Theta }{1-I/MR^{2}}

Answers (1)

As we learnt in

Rolling of a body on an inclined plane -

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

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

- wherein

K=Radius of gyration

\Theta = Angle of inclination

 

 

 a=\frac{mg\ sin\theta}{m+\frac{I}{R^{2}}} =\frac{mg\ sin\theta}{m\left ( 1+\frac{I}{mR^{2}} \right )}=a=\frac{gsin\theta}{1+\frac{I}{mR^{2}}}


Option 1)

\frac{g\sin \Theta }{1-MR^{2}/I}

This is an incorrect option.

Option 2)

\frac{g\sin \Theta }{1+I/MR^{2}}

This is the correct option.

Option 3)

\frac{g\sin \Theta }{1+MR^{2}/I}

This is an incorrect option.

Option 4)

\frac{g\sin \Theta }{1-I/MR^{2}}

This is an incorrect option.

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