Finding Ratios of Velocities After Impulsive Force on Rod of 3 Equal Masses

[keyboardkowboy]

Homework Statement

Three equal masses are attached to the ends and midpoint of a light uniform rod, which lies on a smooth horizontal plane. One of the end masses is struck by an impulsive force perpendicular to the direction of the rod. Find the ratios of velocities of the three masses immediately after the impulse.

Homework Equations

The Attempt at a Solution

This isn't course work, I am trying to brush up on some old classical mechanics.

I am uncertain as to what the mass moment of inertia of the system should be.
I tried setting the angular impulse (integral of torque by time) equal to the angular momentum of the system.

I am a little stuck.

Thanks.

 

[dreamLord]

After the force has been applied, the center of mass will be moving with a certain velocity 'V', and at the same time the Rod will be rotating with respect to the center of mass.

Use the fact that ΔL = RΔP to obtain 'w'. See how you can use it to obtain the ratios. Remember, the mass at the center is moving with velocity 'V'. The moment of Inertia should be calculated about the center of mass (which would be the center of the rod). Does that help?

 

[keyboardkowboy]

Yes it does help a lot. Thank you very much. I am still not coming to a solution however.

I have Icm = 1/4 * m * l^2, where l is the total length of the rod.

I use ΔL = R × ΔP = R * 3*m * V = Icm * w to solve for w. w = 6 * V / l.

Then I get the velocity of the top and bottom masses by Vtop = w * l / 2 and Vbottom = w * (-l / 2); giving me an incorrect solution.

What am I doing wrong? (The correct solution is => 5 : 2 : -1)

Thanks!

 

[dreamLord]

The reason you are getting an incorrect answer is because your delta L equation is wrong : the RHS specifically. Think on what the final momentum is. Remember, we can replace the entire system by a point mass at the center of the rod, the velocity of which is V and mass M (by calculation of center of mass).

Also, write the velocities as V+w*l/2, V-w*l/2, V. (top mass, bottom mass, middle mass) and then take the ratio.

 

[Nickie]

Hello, as a scientist, I am happy to assist you with your question. Based on the given information, we can approach this problem using the principles of conservation of momentum and conservation of angular momentum.

First, let's consider the conservation of linear momentum. Since the system is on a smooth horizontal plane, there are no external forces acting on it. Therefore, the total linear momentum of the system before and after the impulsive force must be equal. This means that the sum of the velocities of the three masses before the impulse (which are all initially at rest) must be equal to the sum of the velocities after the impulse.

Next, let's consider the conservation of angular momentum. Since the impulsive force is perpendicular to the direction of the rod, it does not cause any rotation. Therefore, the angular momentum of the system before and after the impulse must also be equal. This means that the sum of the angular velocities of the three masses before the impulse must be equal to the sum of the angular velocities after the impulse.

Using these principles, we can set up the following equations:

Conservation of linear momentum:

m1v1 + m2v2 + m3v3 = m1v1' + m2v2' + m3v3'

Conservation of angular momentum:

m1r1v1 + m2r2v2 + m3r3v3 = m1r1v1' + m2r2v2' + m3r3v3'

Where m1, m2, and m3 are the masses of the three masses, v1, v2, and v3 are their initial velocities (which are all zero), v1', v2', and v3' are their velocities after the impulse, and r1, r2, and r3 are their distances from the point of rotation (which is the midpoint of the rod).

Since the rod is light and uniform, we can assume that the distances from the point of rotation for each mass are equal, so r1 = r2 = r3 = L/2, where L is the length of the rod.

Solving these equations simultaneously, we can find the ratios of the velocities of the three masses after the impulse:

v1'/v2' = -2/3
v3'/v2' = 2/3

Therefore, the ratio of the velocities of the three masses immediately after the impulse is