Rotation Ques 73
- A solid cylinder rolls without slipping on an inclined plane inclined at an angle $\theta$. Find the linear acceleration of the cylinder. Mass of the cylinder is $M$.
$(2005,4$ M)
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Answer:
Correct Answer: 73.$\frac{2}{3} g \sin \theta \quad$ 29. $a _{CM}=\frac{4 F}{3 m_1+8 m_2}, a _{\text {plank }}=\frac{8 F}{3 m_1+8 m_2}=2 a _{CM}$
(b) $\frac{3 F m_1}{3 m_1+8 m_2}, \frac{F m_1}{3 m_1+8 m_2}$
Solution:
Formula:
- For rolling without slipping, we have
$$ a=R \alpha $$
or $\quad \frac{M g \sin \theta-f}{M}=R \frac{f R}{\frac{1}{2} M R^{2}}$
$$ \begin{aligned} & \text { or } & \frac{M g \sin \theta-f}{M} & =\frac{2 f}{M} \\ & \therefore & f & =\frac{M g \sin \theta}{3} \end{aligned} $$
Therefore, linear acceleration of cylinder,
$$ a=\frac{M g \sin \theta-f}{M}=\frac{2}{3} g \sin \theta $$
BETA
