Rotation Ques 68
- A uniform disc of mass $m$ and radius $R$ is rolling up a rough inclined plane which makes an angle of $30^{\circ}$ with the horizontal. If the coefficients of static and kinetic friction are each equal to $\mu$ and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is and its direction is (write up or down) the inclined plane.
(1997C, 1M)
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Answer:
Correct Answer: 68.$\frac{m g}{6}$, up
Solution:
Formula:
- The equations of motion are
$$ \begin{aligned} a & =\frac{m g \sin \theta-f}{m} \\ & =\frac{m g \sin 30^{\circ}-f}{m}=\frac{g}{2}-\frac{f}{m} \\ \alpha & =\frac{\tau}{I}=\frac{f R}{I}=\frac{f R}{m R^{2} / 2}=\frac{2 f}{m R} \end{aligned} $$
For rolling (no slipping)
$$ \begin{array}{rlrl} a & =R \alpha \\ \text { or } & & g / 2-f / m & =2 f / m \\ \therefore & \frac{3 f}{m}=g / 2 & \text { or } \quad f=m g / 6 \end{array} $$
BETA
