Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

Suppose the Rod with the Balls a and B of the Previous Problem is Clamped at the Centre in Such a Way that It Can Rotate Freely About a Horizontal Axis Through the Clamp. - Physics

Sum

Suppose the rod with the balls A and B of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it. (a) Find the angular momentum and the angular speed of the system just after the collision. (b) What should be the minimum value of h so that the system makes a full rotation after the collision.

Solution

(a) Angular momentum $= m\nu r$

Let the particle P collides the ball B with a speed u and system moves with speed just after the collision.

Applying the law of conservation of linear momentum, we get

$mu = 2m\nu - m\nu = m\nu$

$\therefore u = \nu$

$\text{Velocity, }u = \sqrt{2gh}$

$\text{and }r = \frac{L}{2}$

Initial angular momentum of system about COM of rod,

$mur = m \times \sqrt{2gh} \times \frac{L}{2}$

$= \frac{mL\sqrt{gh}}{\sqrt{2}}$

Angular momentum of system about COM of the rod just after the collision $= l\omega$

$I = \frac{2m L^2}{4} + \frac{m L^2}{4} = \frac{3m L^2}{4}$

Applying the law of conservation of angular momentum and obtaining the value of ω, we get

$\omega = \frac{mL\frac{\sqrt{gh}}{\sqrt{2}}}{\frac{3m L^2}{4}} = \frac{\sqrt{8gh}}{3L}$

(b) When the mass 2m and m are at the top most position and at the lowest point, respectively, they will automatically rotate. In this position, we have

Total gain in potential energy

$= 2mg\left( \frac{L}{2} \right) - mg\left( \frac{L}{2} \right)$

Kinetic energy = $\frac{1}{2}I \omega^2$

Therefore, by the law of conservation of energy, we have

$mg\frac{L}{2} = \frac{1}{2} \times \frac{3}{4}m L^2 \times \left( \frac{8gh}{9g L^2} \right)$

$\Rightarrow h = \frac{3L}{2}$

Is there an error in this question or solution?

APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 64 | Page 199