Linear momentum or Angular Rotation

[Manasan3010]

Homework Statement

A cylinder with mass M and length L is rested on a frictionless surface. A particle with mass m and velocity V comes to rest after an elastic collision with one end of the cylinder. What is the velocity of Center of Gravity of cylinder after collision?

Relevant Equations

-

I think the answer is $\frac{mV}{M}$ but I am not sure. Won't the cylinder tries to rotate due to the collision at one end? Is this anything related to Angular Momentum?

The Answers given were,

 

[Doc Al]

I think the answer is $\frac{mV}{M}$ but I am not sure.

Looks good to me. All you care about is the velocity of the cylinder's center of mass, so conservation of linear momentum is all you need.

Won't the cylinder tries to rotate due to the collision at one end?

Sure, but who cares?

 

[kuruman]

Sure, but who cares?

I think one ought to care enough to investigate the issue. I agree that linear momentum conservation yields the answer $V_{cm}=\dfrac{m}{M}v$. However, how do we know that the problem is not over-determined if one considers the other conservation laws? After all, assuming that the masses are given, after finding $V_{cm}$ there is only one quantity to be determined, the angular speed of the rod, and two conservation equations. Hmm, let's see ...
From angular momentum conservation about the midpoint of the rod
$$mv\frac{L}{2}=\frac{1}{12}mL^2\omega\rightarrow \omega=\frac{v}{L}\left( \frac{6m}{M}\right)$$
From kinetic energy conservation and using the value for $V_{cm}$ found above
$$\frac{1}{2}mv^2=\frac{1}{2}M \left( \frac{m}{M}v\right)^2+\frac{1}{2}\times\frac{1}{12}ML^2\omega^2~\rightarrow~\omega=\frac{v}{L}\sqrt{ \frac{12m(M-m)}{M^2} }$$It follows that the two expressions for $\omega$ are compatible only if $M=4m$. Thus, any future author who rewrites this problem giving specific values for the masses must ensure that the rod is 4 times more massive than the particle to make the solution work. Because the five choices given in this particular question all involve $M$ and $m$ as presumably independent quantities, I am led to believe that the author of this question did not take the constraint on the masses into account. It would have been better to provide the five choices as numerical factors multiplying $v,$ the correct answer being $\frac{1}{4}v$.

 

[Manasan3010]

Logically, Won't the center of gravity(Geometrical center) stay still and the ends would rotate (Like in a fan). If this the case, shouldn't the answer be 0 velocity because of it being in the center?

(Zero motion at the center of fan)

 

[Doc Al]

Logically, Won't the center of gravity(Geometrical center) stay still and the ends would rotate (Like in a fan). If this the case, shouldn't the answer be 0 velocity because of it being in the center?

No, that won't happen. Realize that there is a net force acting on the cylinder during the collision, thus the center of mass will accelerate. (Newton's 2nd law.)

The analysis by @kuruman is well worth your study and places interesting constraints on the relationship between the masses. However, such an analysis is not needed to solve the problem: conservation of linear momentum is enough. (If the author of the problem were on the ball, like @kuruman, he could have asked some interesting followup questions! I suspect he didn't do the analysis.)

 

[kuruman]

Logically, Won't the center of gravity(Geometrical center) stay still and the ends would rotate (Like in a fan). If this the case, shouldn't the answer be 0 velocity because of it being in the center?

Logically, no. In addition to what @Doc Al already pointed out in #5, in post #1 you calculated, and we all agreed, that the speed of the CM after the collision is $V_{cm}=\dfrac{m}{M}v$. To suggest now that it is zero is illogical. Of course, $V_{cm}=0$ in an inertial reference frame moving to the right with velocity $u=\dfrac{m}{M}v$ relative to the lab frame, but the problem (and the offered choices) are relative to the lab frame where the particle is initially moving with speed $v$ to the right and is at rest after the collision.