Rotational Motion

JEE Main level, numerical answer, medium

157 of 280

Question 157 of 280

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of 4ms−14m{s}^{-1}, is _____cmcm.
(take g=10ms−2g=10m{s}^{-2})

Show answer and solution

Answer: 120

The portion of the strings between the ceiling and the cylinder are at rest. Hence, the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings.
Now, from energy conservation
mgh=12mv2+12Iω2mgh=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2}
where, II is moment of inertia of cylinder and ω=vR\omega =\dfrac{v}{R} is angular velocity.
Then, we have
mgh=12mv2+12mR22ω2⇒gh=12v2+12R22(vR)2mgh=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}\dfrac{m{R}^{2}}{2}{\omega }^{2} \Rightarrow gh=\dfrac{1}{2}{v}^{2}+\dfrac{1}{2}\dfrac{{R}^{2}}{2}{\left(\dfrac{v}{R}\right)}^{2}
10h=162+164⇒h=1.2m=120cm10h=\dfrac{16}{2}+\dfrac{16}{4} \Rightarrow h=1.2m=120cm

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Rotational kinematics), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.