Start Free Mock LoginAlready registered? Login here

Physics - Rotation Question with Solution | TestHub

PhysicsRotationTranslational + RotationalHard2 minPYQ_2021

PhysicsHardnumerical

A circular disc reaches from top to bottom of an inclined plane of length $L .$ When it slips down the plane, it takes time $t_{1}$ . When it rolls down the plane, it takes time $t_{2}$ . The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$ . The value of $x$ will be

Answer:

2.00

Solution:

If disk slips on inclined plane, then its acceleration, $a_{1} = g \sin θ$
$L = \frac{1}{2} a_{1} t_{1}^{2}$
$\Rightarrow t_{1} = \sqrt{\frac{2 L}{a_{1}}} \dots (i)$

If disk rolls on inclined plane, its acceleration,
$a_{2} = \frac{g \sin θ}{1 + \frac{I}{m R^{2}}}$
$a_{2} = \frac{g \sin θ}{1 + \frac{m R^{2}}{2 m R^{2}}}$
$a_{2} = \frac{2}{3} g \sin θ$
Now $L = \frac{1}{2} a_{2} \cdot t_{2}^{2}$
$\Rightarrow t_{2} = \sqrt{\frac{2 L}{a_{2}}} \dots (ii)$

Now, $\frac{t_{2}}{t_{1}} = \sqrt{\frac{a_{1}}{a_{2}}} = \sqrt{\frac{3}{2}}$
$\Rightarrow x = 2$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Translational + Rotational

⏱ 2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...