Question 18

NUMERICALHARD

Rotational Dynamics of a Pivoted Circular Disk and Particle Collision

A uniform circular disk of radius $0.2 \text{ m}$ and mass $1 \text{ kg}$ is pivoted at its top point $C$ such that it can rotate freely around $C$ in the $XY$ plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass $20 \text{ g}$ , travelling along negative $x$ direction in the $XY$ plane with speed $100 \text{ ms}^{-1}$ , hits the circumference of the disk at a point $P$ . After collision the particle moves along negative $y$ direction at a speed of $90 \text{ ms}^{-1}$ . [Given: the acceleration due to gravity $(g) = - 10 \hat{j} \text{ ms}^{-2}$ ]

Amount of energy loss (in J) in the collision is:

Correct Answer: 17.5

Detailed Solution

Initial Kinetic Energy: $K_i = \frac{1}{2} m v_1^2 = 0.5 \times 0.02 \times (100)^2 = 100 \text{ J}$ .
Final Kinetic Energy: $K_{f, particle} = \frac{1}{2} m v_2^2 = 0.5 \times 0.02 \times (90)^2 = 81 \text{ J}$ . $K_{f, disk} = \frac{1}{2} I_C \omega^2$ (from previous question calculation) $\approx 1.528 \text{ J}$ . Total $K_f = 81 + 1.528 = 82.528 \text{ J}$ .
Energy Loss: Loss $= K_i - K_f = 100 - 82.528 = 17.472 \text{ J}$ . Rounding to one decimal place, we get $17.5 \text{ J}$ .

Question 18

Rotational Dynamics of a Pivoted Circular Disk and Particle Collision

Detailed Solution

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