Angular momentum + linear momentum =confusion. HElp me?

[vkash]

which one will faster?

consider two cases
case 1: A impulse P is applied to a point object of mass m.velocity of center of mass is v₁

case 2: A impulse P is applied to one end of a rod of mass m length l .velocity of center of mass is v₂(It will rotate also!)

NOTE: all the things are happening in frictional less horizontal plane so g will not matter.

which is greater v₁ or v₂.
I think v₁
because in first case impulse given P is responsible for only change in Linear momentum but in second case applied impulse is responsible for change in linear as well as angular momentum of rod.

I ask this question because there is a question in my book similar to case 2 and they ask velocity of center of mass of rod. If it is point object(as in case 1) then it is definitely $\frac{P}{M}$ but in case of rod i think it should smaller than $\frac{P}{M}$. But how to calculate that?

 

[A.T.]

which is greater v₁ or v₂.

They are the equal.

 

[vkash]

They are the equal.

HOW?
where am i wrong.

 

[rcgldr]

How?

The linear component of acceleration of the center of mass of the rod is equal to the linear component of force / mass, regardless of the point of application of that force. In the case where the force is applied to the end of a rod perpendicular to the direction of P, then initially, there's less inertial resistance to the force, so in order for the force to remain the same, the rate of acceleration at the point of application of the force has to be greater if the inertial resistance is lower. The force ends up being applied at a greater speed and over a longer distance, which results in a greater change in total momentum (linear and angular).

You also have to assume that the direction of P remains constant regardless of the rotation of the rod and that there's no slippage when applying the force. The direction of the center of mass of the rod will be the same as the direction of P. If the impulse P occurs over a long period of time, the motion is complex; during the time the impluse P is being applied, the rod rotates back and forth, including the end point where impulse P is being applied.

 

[vkash]

The linear component of acceleration is equal to force / mass, regardless of the point of application of that force. In the case where the force is applied to the end of a rod perpendicular to the direction of P, then initially, there's less inertial resistance to the force, so in order for the force to remain the same, the rate of acceleration at the point of application of the force has to be greater if the inertial resistance is lower. The force ends up being applied at a greater speed and over a longer distance, which results in a greater change in total momentum (linear and angular).

You also have to assume that the direction of P remains constant regardless of the rotation of the rod and that there's no slippage when applying the force. If the impulse is strong (force) enough and long (time) enough, the rod doesn't really rotate, but instead will swing back and forth like a pendulum with the pivot point at the point of application of the impulse P on the end of the rod, during the time that the impulse P is being applied.

Does not understand fully. what is angular momentum of rod and velocity of center of mass after impulse has been imparted. Impulse is imparted perpendicularly on the end of rod.
Can you please solve this for me.

 

[rcgldr]

Does not understand fully.

I just udpated my previous post. Assuming the direction of P is constant, the direction of the center of mass of the rod is the same as the direction of P. If the impulse P occurs over a long enough period of time, the rod rotates back and forth, including the end of the rod where the impulse P is being applied.

As mentioned by A. T., the velocity of the center of mass of the rod is the same as the velocity of a point mass after impulse P is applied. It might help to visualize this by considering the direction of P to be horizontal. Since there's no vertical force, there's no vertical acceleration of the center of mass. In the case of the rod, the point of application of P (but not it's direction) will be moving vertically over time (since the center of mass is moving horizontally while the rod rotates back and forth).