Find The Percent Change In K.E. Of Rod Hit By Putty

• XanMan
In summary, the conversation discusses a problem involving a rod and a piece of putty colliding on a frictionless surface, and the task of finding the percentage of kinetic energy lost during the collision. The solution involves calculating the velocity and angular momentum before and after the collision, and using the conservation of angular momentum to determine the answer. The final result is a 20% loss of kinetic energy during the collision.
XanMan
I'm trying to solve this problem, however I'm unsure if my reasoning is correct - I do agree with the answer, but something seems wrong to me in my reasoning.

1. Homework Statement

A rod is on a frictionless surface, and is defined by the length L and mass M. It is hit on one end by a piece of putty, also of mass M, with a velocity V. The putty sticks to the rod and there is an inelastic collision between the two (of very short duration). What is the percentage of Kinetic Energy is lost during the collision?

The question has 3 more parts before the percentage bit. I shall highlight them briefly:
i) Find the velocity of the CoM before and after the collision: ##\frac{V}{2}##
ii) Find the angular momentum about the CoM before the collision: ##L = \frac{MVL}{4}##
iii) Find the angular velocity about the CoM after the collision: ##\omega = \frac{6V}{5L}##

I worked out iii) using the conservation of angular momentum, which I am unsure if it is the correct way. The rod has no form of pivot, it will rotate about its centre of mass. However it will also have linear motion. Due to the inelastic collision, I know that linear momentum is conserved. However, is angular momentum conserved?

Homework Equations

[/B]
I found the CoM to be located at ##\frac{3L}{4}##
$$Total KE = Linear KE + Rotational KE$$
$$Linear KE = \frac{MV^2}{2}$$
$$Rotational KE = \frac{I \omega ^2}{2} = \frac{L \omega}{2}$$

The Attempt at a Solution

[/B]
Before collision there is no Rotational KE. Moreover, only the putty is in motion in the system. Thus the KE Before is found using the mass of the putty and its velocity.

$$KE_{Before} = \frac{MV^2}{2}$$

The KE After is a combination of Linear and Rotational KE. Let's work them out in two separate parts.

a) Linear KE:

The velocity of the CoM after collision is ##\frac{V}{2}## and the combined mass is ##2M##. Thus:

$$Linear KE = \frac{MV^2}{4}$$

b) Rotational KE:

We must first find the moment of inertia about the CoM. The moment of inertia of the system about the other end of the rod is:

$$I = I_{Rod} + I_{Putty} = \frac{ML^2}{3} + ML^2 = \frac{4ML^2}{3}$$

By Parallel Axis Theorem:

$$I_{CoM} = I - M_T X_{CoM}^2 = \frac{4ML^2}{3} - \frac{18ML^2}{16} = \frac{5ML^2}{24}$$

Finding the Rotational KE:

$$Rotational KE = \frac{I \omega ^2}{2} = \frac{3MV^2}{20}$$

Thus:

$$KE_{After} = \frac{MV^2}{4} + \frac{3MV^2}{24} = \frac{2MV^2}{5}$$

We're almost there! Huzzah!

By finding the difference between the KE Before and KE After, one sees that the percentage difference is 20%.

Last edited by a moderator:
XanMan said:
However, is angular momentum conserved?
Definitely. There are no external torques, so angular momentum about the CoM will be conserved.

Your solution looks good to me.

XanMan
Doc Al said:
Definitely. There are no external torques, so angular momentum about the CoM will be conserved.

Your solution looks good to me.

Thanks Doc Al! All the best for the new year.

1. What is the formula for calculating percent change in kinetic energy?

The formula for calculating percent change in kinetic energy is: (Final kinetic energy - Initial kinetic energy) / Initial kinetic energy x 100%.

2. How do you find the final kinetic energy of a rod hit by putty?

To find the final kinetic energy, you will need to know the mass, velocity, and height of the rod before and after it was hit by putty. Use the formula: KE = 0.5 x mass x velocity^2 + mass x gravity x height. Plug in the values for both before and after the putty hit and subtract the initial kinetic energy from the final kinetic energy.

3. Can the percent change in kinetic energy be negative?

Yes, the percent change in kinetic energy can be negative if the final kinetic energy is less than the initial kinetic energy. This means that the putty caused a decrease in the rod's kinetic energy.

4. How does the mass of the putty affect the percent change in kinetic energy of the rod?

The mass of the putty will directly affect the percent change in kinetic energy of the rod. The more mass the putty has, the more kinetic energy it will transfer to the rod, resulting in a larger percent change.

5. What factors can affect the accuracy of the calculation for percent change in kinetic energy?

The accuracy of the calculation for percent change in kinetic energy can be affected by factors such as measurement errors, external forces acting on the rod, and the assumption that all energy transfer occurs between the putty and the rod. It is important to carefully measure and consider all factors when calculating the percent change in kinetic energy.

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