Calculating Angular Momentum: Uniform Rod and Point Mass Collision

[avenkat0]

Homework Statement

A uniform rod (M 4.4 kg, L 0.827 m) starts at rest on a frictionless table. A point mass = 0.808 kg hits the rod at a right angle at speed 6.39 m/s. The block strikes the rod at a distance of 0.12 m below the center of mass, and stops. Assume the block does not stick to the rod. Find:

1. vf, the speed of the center of mass of the rod after the collision.

2. Find the energy lost?

Homework Equations

The Attempt at a Solution

Vf of the cm...
i got omega by finding the linear momentum imparted upon the rod by the force of the colliding object and dividing it by the moment of inertia of the rod rotationg about the center of mass...
but now for the V of the cm... if the rod were just rotating it would be zero but the table is frictionless so there is translational motion...
so i used the formula... .5Mblock*Vblock^2=.5Mrod*Vrod^2 + .5iomega^2
and Vrod came out to be wrong...

is there a flaw in my reasoning...? thank you

 

[borgwal]

Vf of the cm...
i got omega by finding the linear momentum imparted upon the rod by the force of the colliding object and dividing it by the moment of inertia of the rod rotationg about the center of mass...

You go wrong right at the first step. if you insist on calculating omega first, you didn't calculate the angular momentum of the colliding object before the collision.

Fortunately for you the question doesn't ask for omega, but only for the cm velocity. That's easy to calculate: use conservation of...(not energy!)

 

[thomate1]

for your time

I would like to clarify that the question is asking for the speed of the center of mass of the rod after the collision, not the angular velocity. The formula for angular momentum is not applicable in this scenario.

To solve this problem, we can use the conservation of momentum and energy principles.

1. To find the speed of the center of mass of the rod after the collision, we can use the conservation of momentum equation:

m1v1 + m2v2 = (m1+m2)vf

Where m1 and v1 represent the mass and velocity of the point mass, m2 and v2 represent the mass and velocity of the rod, and vf represents the final velocity of the center of mass after the collision.

Substituting the given values, we get:

(0.808 kg)(6.39 m/s) + (4.4 kg)(0 m/s) = (0.808 kg + 4.4 kg)vf

Solving for vf, we get vf = 1.17 m/s.

2. To find the energy lost during the collision, we can use the conservation of energy equation:

KE1 + PE1 = KE2 + PE2

Where KE1 and PE1 represent the initial kinetic and potential energies, and KE2 and PE2 represent the final kinetic and potential energies.

Initially, the rod and point mass have no kinetic energy and only potential energy due to the height of the point mass above the ground. After the collision, the point mass stops and the rod has a final kinetic energy, but no potential energy.

Substituting the given values, we get:

0 + (0.808 kg)(9.8 m/s^2)(0.12 m) = (0.808 kg)(0 m/s^2) + (4.4 kg)(1.17 m/s)^2

Solving for the energy lost, we get 2.17 J.

In conclusion, the speed of the center of mass of the rod after the collision is 1.17 m/s and the energy lost during the collision is 2.17 J.