JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Rotational Mechanics

A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point

C

such that it can rotate freely around

C

in the

X Y

plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative

x

direction in the

X Y

plane with speed

100 ms^{- 1}

, hits the circumference of the disk at a point

P

. After collision the particle moves along negative

y

direction at a speed of

90 ms^{- 1}

. [Given: the acceleration due to gravity

(g) = - 10 \hat{j} ms^{- 2}

]

After the collision the disk starts to rotate around point

C

in the

X Y

plane. The maximum change in the height (in m) of its center

O

is:

Answer: 0.15

Solution

Conservation of angular momentum about

C

(R + \frac{R}{2}) m v_{i} = I ω + \frac{R}{2} m v_{f}

.

With

I = \frac{3}{2} M R^{2}

(disk pivoted at edge),

m = 0.02

kg,

M = 1

kg,

R = 0.2

(1 + \frac{1}{2}) \times 0.02 \times 100 = \frac{3}{2} \times 1 \times (0.2)^{2} \times ω + \frac{0.2}{2} \times 0.02 \times 90

.

Solving:

ω \approx 7.138

rad/s.

Energy conservation:

M g R (1 - cos θ) = \frac{1}{2} I ω^{2}

1 \times 10 \times 0.2 \times (1 - cos θ) = \frac{1}{2} \times \frac{3}{2} \times 1 \times (0.2)^{2} \times (7.138)^{2}

.

Solving:

R (1 - cos θ) \approx 0.15

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer