Question

A disc of 2kg and radius 10 cm rotates about its vertical axis of symmetry with an angular speed 10 rad/sec. A point mass 1kg is thrown at velocity $V_0$ towards disc. The vertical distance (b) of mass from centre of disc rotation is 8 cm. The point like mass sticks to disc.

7. If disc rotates such that its axis remains at rest than find value of $V_0$ for which the system stops rotating after collision.

• A. $\frac{15}{8}$m/s
• B. $\frac{17}{3}$m/s
• C. $\frac{7}{8}$m/s
• D. $\frac{23}{8}$m/s

Answer 1

5/4 m/s

Answer 2

The problem involves the conservation of angular momentum during an inelastic collision. We need to calculate the initial angular momentum of the disc and the incoming point mass, and then equate their magnitudes for the system to stop rotating.

1. Calculate the initial angular momentum of the disc ($L_{disc}$):

The moment of inertia of a disc about its axis of symmetry is $I_{disc} = \frac{1}{2}MR^2$. $I_{disc} = \frac{1}{2}(2 \text{ kg})(0.1 \text{ m})^2 = 0.01 \text{ kg m}^2$. The initial angular momentum of the disc is $L_{disc} = I_{disc}\omega_0$. $L_{disc} = (0.01 \text{ kg m}^2)(10 \text{ rad/s}) = 0.1 \text{ kg m}^2\text{/s}$.

2. Calculate the initial angular momentum of the point mass ($L_{mass}$):

The angular momentum of a point mass about an axis is given by $L_{mass} = m V_0 b$, where 'b' is the perpendicular distance from the axis to the line of action of $V_0$. $L_{mass} = (1 \text{ kg}) V_0 (0.08 \text{ m}) = 0.08 V_0 \text{ kg m}^2\text{/s}$.

3. Apply conservation of angular momentum:

For the system to stop rotating after the collision, the final angular momentum must be zero. This means the initial angular momentum of the disc and the point mass must be equal in magnitude and opposite in direction. $L_{disc} = L_{mass}$ $0.1 = 0.08 V_0$ $V_0 = \frac{0.1}{0.08} = \frac{10}{8} = \frac{5}{4} \text{ m/s} = 1.25 \text{ m/s}$.