# [SOLVED] Mechanics question

• February 25th 2009, 12:49 PM
jackiemoon
[SOLVED] Mechanics question
Q. A rod of mass m and length l can rotate freely around the fixed pivot point A (see picture attached) and is initially at an angle θ = 60 degrees with the vertical direction. The rod swings under the effect of gravity and its end point hits a point-like mass m (equal to that of the rod) initially at rest. Compute the maximum angle of oscillation of the rod after the collision a) if the collision is elastic b) if the collision is completely inelastic ie. if the particle sticks to the rod.

Don't know where to start with this one. Any help would be very much appreciated.
• February 26th 2009, 12:54 AM
Mechanics
Hello jackiemoon
Quote:

Originally Posted by jackiemoon
Q. A rod of mass m and length l can rotate freely around the fixed pivot point A (see picture attached) and is initially at an angle θ = 60 degrees with the vertical direction. The rod swings under the effect of gravity and its end point hits a point-like mass m (equal to that of the rod) initially at rest. Compute the maximum angle of oscillation of the rod after the collision a) if the collision is elastic b) if the collision is completely inelastic ie. if the particle sticks to the rod.

Don't know where to start with this one. Any help would be very much appreciated.

This is what you need:

1 The moment of inertia of the rod about A, $I = \tfrac{1}{3}ml^2$

2 The centre of mass of the rod is initially at a depth $\tfrac{1}{2}l\cos 60^o = \tfrac{1}{4}l$ below A

3 At the moment of impact the centre of mass is at a depth $\tfrac{1}{2}l$ below A.

4 Use the energy principle to find the KE ( $= \tfrac{1}{2}I\omega^2$) of the rod just before impact, where $\omega$ is its angular velocity.

5 At the moment of impact the only external impulse on the system (rod and particle) is at A. So the angular momentum of the system about A is conserved. (Angular momentum of rod = $I\omega$ before impact and $I\omega'$ after impact where $\omega'$ is the angular velocity of the rod after impact; angular momentum of particle = $0$ before impact and $mvl$ after impact, where $v$ is the initial speed of the particle.)

Then:

6 If the collision is perfectly elastic, there is no loss of energy. So you can equate the KE of the system before and after impact.

7 If the collision is perfectly inelastic, and the particle adheres to the rod then there will be loss of energy (so you can't use #6). But the speed, $v$, of the particle will now be $l\omega'$ = linear speed of the end of the rod.

Can you attempt it now?

• February 26th 2009, 04:45 AM
skeeter

for jackiemoon ...

re: the perfectly inelastic collision, don't forget that the moment of inertia changes when the point mass sticks to the rod

$
I_f = I_{rod} + I_{mass}
$
• February 26th 2009, 05:41 AM
Quote:

Originally Posted by skeeter
re: the perfectly inelastic collision, don't forget that the moment of inertia changes when the point mass sticks to the rod

$
I_f = I_{rod} + I_{mass}
$

I don't want to confuse things here, but you have two choices as far as the system's angular momentum about A is concerned, after the impact:

• You can say that after the impact, you now have a new single body comprising the rod with the particle joined to it. If you do it this way, find the new moment of inertia $I_f$ as skeeter has indicated above. And so the angular momentum of the new body will be $I_f\omega'$.
• Or you can treat the rod and the particle as separate bodies that are moving as one, and say (as I did in my first posting) that the angular momentum of the particle about A is $mvl$. (Why? Because its angular momentum about A is defined as the moment about A of its linear momentum $= mv \times l$.) You'll then add this to the angular momentum of the rod, $I\omega'$, to give the total angular momentum of the system.

Either way will give the same result.