System Of Particles And Rotational Motion Ques 77
77. A solid cylinder of mass $2 $ $kg$ and radius $50$ $ cm$ rolls up an inclined plane of angle of inclination $30^{\circ}$. The centre of mass of the cylinder has speed of $4$ $ m / s$. The distance travelled by the cylinder on the inclined surface will be, [take $g=10$ $ m / s^{2}$ ]
[NEET Odisha 2019]
(a) $2.4 $ $m$
(b) $2.2 $ $m$
(c) $1.6 $ $m$
(d) $1.2 $ $m$
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Answer:
Correct Answer: 77.(a)
Solution:
- (a) Kinetic energy at the bottom
$=\frac{1}{2} m v_0^{2}+\frac{1}{2} I \omega^{2}=\frac{3}{4} m v_0^{2}$
As $I _{\text{solid cylinder }}=\frac{1}{2} MR^{2}$ and $V=R \omega$
From work - energy theorem
$ \begin{aligned} & \frac{3}{4} m v_0^{2}=m g h=m g d \sin \theta \\ d & =\frac{\frac{3}{4} mv_0^{2}}{mg \sin \theta} \\ \Rightarrow \quad & d=\frac{3}{4} \frac{v_0^{2}}{g \sin \theta}=\frac{3}{4} \cdot \frac{16}{10 \times 1 / 2}=2.4 m \end{aligned} $
BETA
