Physics problem on angular momentum

Question

Problem.

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10 m/s The ball rebounds in the opposite direction with a speed of 6 m/s (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

My attempt.

I've tried $mv = I \omega$ but that doesn't work. I've tried to see the relationship between momentum but I'm just missing something. Can someone clarify exactly what is going on? I don't have the radius of the ball so I can't set $I \omega_{ball} = I \omega_{rod}$...

P.S. I know this is a physics question but there is a physics tag and I like mathematics SE a lot better than physics SE.

Answer 1

When dealing with angular momentum issues you need to choose your reference point appropriately.

Considering the rod and ball as our system, the only external forces are gravity and any force exerted by the frictionless hinge. There is no external torque applied, so angular momentum is conserved.

We need to pick a suitable reference point to compute angular momentum. Since the hinge applies a force to the rod and I don't want to bother computing this force explicitly, we will take the hinge as the reference point (hence the force on the hinge contributes nothing to the angular momentum).

The moment of inertia of a rod of length $L_{rod}$ of mass $m$ about an end point is $I = {m L_{rod}^2 \over 3}$.

The ball imparts an impulsive change of momentum. You can model this as a distribution (impulse 'function') or as the limit of a large force over a short time, but the net result is that there is an instantaneous change of momentum $P=3 \cdot 16 \text{ kg }m/s$ applied at a distance $L_{action}=1.5 m$ below the hinge.

This translates to an instantaneous change of angular momentum (of the rod) of $Q= L_{action} P$ (in a horizontal direction).

Let $\theta$ be the angle of the rod from vertical. Initially $\theta = 0$, and the tangent of any motion at $\theta = 0$ is horizontal and hence perpendicular to gravity. Consequently gravitational forces contribute nothing (instantaneously, at least).

The angular acceleration satisfied $I \ddot{\theta} = T$, where $T$ is the applied torque, in this case we are dealing with an impulsive change so we have (integrating over a small time period and taking limits, if you wish) $I (\dot{\theta}(0) - 0) = Q$, or in other words, $\dot{\theta}(0) = {Q \over I}$.

Now substitute all quantities above to get $\dot{\theta}(0) = { L_{action} P \over {m L_{rod}^2 \over 3}}$, note that the mass of the rod is $m = {90 \over g} kg$, where $g \approx 9.81 m/s^2$. This gives $\dot{\theta}(0) \approx 5.886 \,rad/s$.

For part (b), you can either compute the total angular momentum before and after (remember to include the angular momentum of the ball about the hinge just after the bounce), or note that there is no external torque applied to our system. There is an external (horizontal) force applied at the hinge, otherwise the rod would slide and not swing.

Finally, note that this problem is 'nice' because gravity is nicely orthogonal to the initial (tangent to) motion and the hinge force contributes nothing to the angular momentum because of our choice of reference point.

Answer 2

I think when it says "small ball", you may assume that it is a "point ball" (of negligible radius).

I don't know exactly which symbols you use, but you should be able to do it with conservation of energy and/or angular momentum (provided you explain why the latter holds, as in part ($b$)).

By determining the change in momentum of the ball (remembering the change in direction), this will give you the impulse at the point of collision (ie $1.5m$ from the ceiling). Then you can use moments (torque) about the point on the ceiling. The metal bar is uniform and thin, so assume that its centre of mass is at the centre (ie $1m$ from the ceiling).

For part ($b$), consider the fact that there is rotation of the rod about a fixed point. This fixed point will be exerting a force on the rod. Use this to show the conservation or lack of in each case.

Hope this helps! :)

Answer 3

Conservation of Angular Momentum about the hinge:

Let $v_0$ be the initial velocity of the ball,

Let $v_f$ be the final velocity of of the ball

Using conservation of angular momentum about hinge and \omega being the angular velocity of the rod, we get

The distance "l" between the hinge to the impinging point = 1.5 m

Thus, $$lm_{1}v_{0} = -lm_{1}v_{f} + I.\omega$$

$$ I = \frac{1}{3}\dfrac{90}{9.81}2^{2} = 12.2324$$, where L = 2m

$$ 1.5*3*(10+6)= 12.2324.\omega$$

$$\omega = 5.886 rad/sec$$