Rotational Motion

JEE Main level, numerical answer, medium

251 of 280

Question 251 of 280

A solid cylinder is released from rest from the top of an inclined plane of inclination 30∘30^{\circ} and length 60cm60cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ ms−1m{s}^{-1}. (Given g=10ms−2g=10m{s}^{-2})

Show answer and solution

Answer: 2

Let speed of cylinder upon reaching the bottom of the inclined plane be vv.
Here, gain in kinetic energy==loss in potential energy
So, 12Iω2+12mv2=mglsin⁡θ\dfrac{1}{2}I{\omega }^{2}+\dfrac{1}{2}m{v}^{2}=mgl\sin \theta
⇒12(mr22)ω2+12mv2=mglsin⁡30∘\Rightarrow \dfrac{1}{2}\left(\dfrac{m{r}^{2}}{2}\right){\omega }^{2}+\dfrac{1}{2}m{v}^{2}=mgl\sin 30^{\circ}
For pure rolling, v=rωv=r\omega
⇒12(mr22)(vr)2+12mv2=mglsin⁡30∘\Rightarrow \dfrac{1}{2}\left(\dfrac{m{r}^{2}}{2}\right){\left(\dfrac{v}{r}\right)}^{2}+\dfrac{1}{2}m{v}^{2}=mgl\sin 30^{\circ}
⇒14mv2+12mv2=mglsin⁡30∘⇒34mv2=mgl2⇒v=4gl6\Rightarrow \dfrac{1}{4}m{v}^{2}+\dfrac{1}{2}m{v}^{2}=mgl\sin 30^{\circ} \Rightarrow \dfrac{3}{4}m{v}^{2}=\dfrac{mgl}{2} \Rightarrow v=\sqrt{\dfrac{4gl}{6}}
Putting the values, we have
⇒v=4×10×0.66=2ms−1\Rightarrow v=\sqrt{\dfrac{4\times 10\times 0.6}{6}}=2m{s}^{-1}

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 (Rolling without Slipping), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.