# Sticky Collision with Clay Ball and Rod

1. Dec 26, 2007

### Squashy

[SOLVED] Sticky Collision with Clay Ball and Rod

The problem statement, all variables and given/known data

A uniform rod of mass m and length $$\ell$$ = 2 meters is suspended from one end by a frictionless pivot so that it can swing freely in the x-y plane. When the rod is at rest it is struck by a clay ball of equal mass m with initial velocity $$v_0$$ = 10 m/s at an angle of 60º with the vertical rod. The clay ball strikes the rod at the free end and sticks to it. What is the linear velocity of the free end just after the collision?

• 11.4 m/s
• 6.50 m/s
• 3.75 m/s
• 1.85 m/s
• 1.65 m/s

Relevant equations

The moment of inertia of the thin rod about one end is $$I=\frac{1}{3} m L^2$$.

My Background

My Physics C class hasn't started rotational motion/torque yet, and I'm not too comfortable with the equations. (We just finished linear momentum.) This question is a bonus question on our holiday homework, so I know I should figure it out myself, but I've been reading through my textbook and on websites for hours. This is a last resort.

What I've managed to figure out (I think)

The moment of inertia at the end of the rod after the clay ball hits will be $$I=\frac{1}{3} m \ell^2+m \ell^2 = \frac{4}{3} m \ell^2$$, if I'm correct in thinking that moments of inertia can be added like that.

I also figure that the answer can't be A, because it's greater than the clay ball's initial velocity. The clay ball's velocity perpendicular to the rod (i.e. the velocity that will affect the rod's rotation) is 10sin60º = 8.66 m/s. Since choices B through E are less than that, I figure that they are all possibilities.

Attempt 1: Rotational Newton's Second Law

I was thinking that I could do something with torque or a rotational analogue of Newton's Second Law, but it seems that to figure out the force the clay ball exerts on the rod, I'd need to know the time for which it was exerting the force. That isn't provided. I think I'd also need to know the change in momentum, but if I knew that, then I'd know the velocity of the clay ball + rod system on impact, and I'd have the problem solved. So I don't think torque or impulse equations will work.

Attempt 2: Kinetic Energy

\begin{align*} KE_i=KE_f\\ \frac{1}{2} m {v_0}^2=\frac{1}{2} I \omega^2\\ m {v_0}^2=\frac{4}{3} m \ell^2 \omega^2\\ {v_0}^2=\frac{4}{3} \ell^2 \omega^2 \end{align*}

I plugged in 10sin60º for $$v_0$$, because even though it isn't the true initial velocity of the clay ball, I thought it would be the only velocity component affecting the motion of the clay ball + rod system. I figured that the KE from the velocity component parallel to the rod would be dissipated as heat. (I'm probably treating KE too much like a vector.) So, solving for $$\omega$$ and using the equation $$\omega=v r=v \ell$$, I got v = 7.50 m/s, which isn't too close to any of the choices.

Attempt 3: Conservation of Momentum

\begin{align*} P_i=P_f\\ mv_0=I\omega\\ mv_0=\frac{4}{3}m\ell^2vr\\ v_0=\frac{4}{3}\ell^3v\\ \end{align*}

According to Wikipedia, "for an object with a fixed mass that is rotating about a fixed symmetry axis," angular momentum = $$I\omega$$. I wasn't really sure if that applied to this situation. I'm also pretty sure that I can't equate linear momentum and angular momentum like that, but hey, I thought I'd try. So, plugging in 10sin60º for $$v_0$$ again, I obtained a value of v = .812 m/s. Way off from my other answer, and again, not close to the choices.

Help!

I think my main problem, aside from not knowing the formulas well enough, is that I don't know how to deal with the fact that the clay ball is colliding with the rod at a 60º angle rather than perpendicularly. I think I could make a kinetic energy formula or some sort of a momentum formula would work if it were a perpendicular collision. Anyway, if someone could point me in the right direction (and maybe point out my mistakes), I'd really appreciate it!

Last edited: Dec 26, 2007
2. Dec 26, 2007

### HallsofIvy

Staff Emeritus
What I would do is, at least at first, ignore the fact that the rod is attached at one end- or that it is a rod at all. Treat both clay ball and rod as though they were "point masses" and use conservation of momentum to determine the velocity vector of the rod-clay ball combination immediately after the collision as if they were point masses. Then break that into components parallel to and perependicular to the rod. Since the rod is prevented from moving accept around its pivot, the resultant velocity vector will be the component perpendicular to the rod.

3. Dec 26, 2007

### Staff: Mentor

Good!

Since it's an inelastic collision, KE will not be conserved--so forget about that approach.

Also, linear momentum is not conserved (since the rod is constrained at one end)--so forget about that.

Conservation of angular momentum is the way to go.
You want to set initial angular momentum equal to final angular momentum. Here you set the initial linear momentum equal to the final angular momentum. No good. (The units won't even match.)

Read this: Angular Momentum of a Particle

So correct the left hand side of your initial equation and do over. (Done correctly, the initial angular momentum will involve the length of the rod and the sine of the angle.)

Two other errors:
(1) $\omega \ne vr$! ($\omega = v/r$, where $r = \ell$, of course.)
(2) You need to solve for the final speed, not the initial speed!

4. Dec 26, 2007

### Squashy

Thank you for the responses!

I didn't realize that objects moving linearly could have angular momentum; I thought angular momentum only applied to objects undergoing rotational motion. Now I see that all objects in motion have angular momentum, with respect to an origin. Thanks!

So, changing my equation, I get

\begin{align*} L_i=L_f\\ mvr\sin\theta=I\omega\\ 10\sin{60}^\texttt{o}m\ell=\frac{4}{3}m\ell^2\frac{v}{\ell}\\ v=\frac{3}{4}(10\sin{60}^\texttt{o})\\ v=6.50\: \texttt{m/s}\\ \end{align*}

Choice (B).

Those errors were a result of tired brain. Thank you for correcting them.

Last edited: Dec 26, 2007