Physics · Rotational Dynamics

JEE Advanced 2022 — Paper 1 — Question 24

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination $\theta=30^{\circ}$ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance $\mathrm{r}=0.5 \mathrm{~m}$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is $\qquad$ $\mathrm{ms}^{-2}$ . (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ .)

Question figure

Answer: 2.85

Numerical answer — enter this value.

Step-by-step solution

$m g \sin \theta \times R+1 \times \frac{R}{2}-1 \times \frac{3 R}{2}=I_{P} \alpha$

$\frac{10}{2} \times 1+\frac{1}{2}-\frac{3}{2}=\frac{7}{5} \times \mathrm{mR}^{2} \alpha$

$5-1=\frac{7}{3} \alpha$

$\frac{20}{7}=\alpha$

$\mathrm{a}_{\mathrm{cm}}=\mathrm{R} \alpha=\frac{20}{7}$

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion

At time t=0 , a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of alpha=2/3 rad s…

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