JEE Main 2025 Jan 24 Shift 1, Physics Q11: Rolling Motion

JEE Main2025Jan 24, Shift 1Physics

A uniform solid cylinder of mass ' $m$ ' and radius ' $r$ ' rolls along an inclined rough plane of inclination $4 5^{\circ}$ . If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-

A
$\frac{1}{2} g$
B
$\frac{1}{3 2} g$
C
$\frac{2 g}{3}$
D
$2 g$

Solution

For a body rolling without slipping down an incline of angle $θ$ , the standard result is

$a = \frac{g sin θ}{1 + \frac{I}{m r ^{2}}}$

For a uniform solid cylinder, $I = \frac{1}{2} m r^{2}$ , so $\frac{I}{m r ^{2}} = \frac{1}{2}$ and the denominator is $\frac{3}{2}$ .

With $θ = 4 5^{\circ}$ , $sin θ = \frac{1}{2}$ :

$a = \frac{g / 2}{3/2} = \frac{2 g}{3 2} = \frac{2 g}{3}$

$∴$ Option (3).

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JEE Main 2025 Jan 24 Shift 1, Physics Q11: Rolling Motion

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