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  • A circular disc of mass is rolling on inclined plane of inclination <img alt="\mathrm{\theta}" src="h

A circular disc of mass \mathrm{M} is rolling on inclined plane of inclination \mathrm{\theta}, what's friction force on the disc.
 

Option: 1

\mathrm{m g \sin \theta}

 


Option: 2

\mathrm{\frac{1}{2} m g \sin \theta}
 


Option: 3

\mathrm{2 m g \sin \theta}
 


Option: 4

\mathrm{\frac{1}{3} m g\sin \theta }


Answers (1)

best_answer

\mathrm{a=\frac{g \sin \theta}{1+\frac{I_c}{M R^2}}}

For circular disc \mathrm{I_c=\frac{1}{2} mR^{2}}

\mathrm{ a=\frac{g \sin \theta}{1+\frac{1}{2} m R^2 \times \frac{1}{m R^2}} }

\mathrm{ a=\frac{2 g \sin \theta}{3}}                    ..........(1)

To calculate frictional force \mathrm{f_\gamma=\frac{I \cdot a}{R^2}}

\mathrm{ f_\gamma=\frac{m R^2}{2} \times \frac{2}{3} g \sin \theta \times \frac{1}{R^2} }

\mathrm{ f_\gamma=\frac{1}{3} m g \sin \theta}

Hence option 4 is correct.
 

Posted by

mansi

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