Rotation 3 Question 7
13. Statement I Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
Statement II By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
$(2008,3$ M)
Passage Based Questions
Passage 1
A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk diammetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$.
The disk is initially at its equilibrium position with its centre of mass (CM) at a distance $L$ from the wall. The disk rolls without slipping with velocity $\mathbf{v} _0=v _0 \hat{\mathbf{i}}$. The coefficient of friction is $\mu$.
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Answer:
Correct Answer: 13. (d)
Solution:
- In case of pure rolling on inclined plane,
$$ a=\frac{g \sin \theta}{1+I / m R^{2}} $$
$$ \begin{array}{ll} \therefore \quad & I _{\text {solid }}<I _{\text {hollow }} \\ & a _{\text {solid }}>a _{\text {hollow }} \end{array} $$
$\therefore$ Solid cylinder will reach the bottom first. Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant.
$\therefore(KE) _{\text {Hollow }}=(KE) _{\text {Solid }}=$ decrease in PE $=m g h$
$\therefore$ Correct option is $(d)$.
