# Acrobat grabbing monkey

1. Dec 10, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
A circus acrobat of mass $M$ leaps straight up with initial velocity $v_0$ from a trampoline. As he rises up, he takes a trained monkey of mass $m$ off a perch at height $h$, above the trampoline. What is the maximum height attained by the pair?

2. Relevant equations
center of mass

3. The attempt at a solution

I put that $y_1(t)$ is the trajectory of the acrobat, $y_2(t)$ the trajectory of the monkey, and $t_e$ grabbing time. I look at the system acrobat-monkey before and after grabbing.

Before grabbing: Between $0\le t\le t_e$ :
1. $y_1(t) = tv_0 - \frac{gt^2}{2}$
2. $y_2(t) = h$
so the vertical component of the center of mass is
$R_y^{(b)}(t) = \frac{1}{m+M} (my_2+My_1) = \frac{1}{m+M} (mh+Mt(v_0 - \frac{gt}{2}) )$

After grabbing: When $t_e \le t$, the center of mass is under external force $-(m+M)g\vec j$, and its motion is given by:
$R_y^{(a)}(t) = h + (t-t_e) v_1 - \frac{g}{2} (t-t_e)^2$

At maximum height, we will have $\dot R_y^{(a)}(t_{\max}) = 0$ so that $t_{\max}- t_e = \frac{v_1}{g}$.
The maximum height will be $R_y^{(a)}(t_{\max}) = h + \frac{v_1^2}{2g}$

Now we must find $v_1$.
I assume continuity of the center of mass speed at time $t_e$ ( I'm not sure about that! ), and get that $v_1 = \frac{M}{m+M}(v_0 - g t_e)$ where the parenthesized term is the speed of the acrobat at grabbing time.

To conclude I must get rid of $t_e$ which is not a constant of the problem. How should I do that please?

2. Dec 10, 2014

### BvU

When grabbing, y1(te) = h too.

However, there might be a much simpler way to deal with this exercise...

3. Dec 10, 2014

### geoffrey159

Hi,
So you solve the equation $\frac{g}{2} t_e^2 -v_0 t_e + h = 0$ ? You probably keep the first root since the pair will fall back to height $h$, therefore $t_e = \frac{v_0 - \sqrt{v_0^2 - 2 hg}}{g}$

and maximum height is

$(1 - {(\frac{M}{m+M})}^2) h + {(\frac{M}{m+M})}^2 \frac{v_0^2}{2g}$

Correct?

Last edited: Dec 10, 2014
4. Dec 10, 2014

### BvU

I get that too.

And my idea of a simpler way doesn't apply: the grabbing process is an inelastic collision, so there is no conservation for kinetic + potential energy.
There is momentum conservation (your continuity of c.o.m. speed) that in fact demonstrates this non-conservation.
Nice exercise!

5. Dec 10, 2014

### geoffrey159

Thanks BvU !
I am a starter in mechanics so I have no knowledge of what is energy for now.
I am interested by your explanation about continuity of c.o.m speed which I assumed but not quite well understood.
You justified it by momentum conservation which I understand it prooves continuity. But why is momentum conserved around time $t_e$ ?

6. Dec 10, 2014

### BvU

Yes. There is no external force doing work on the system man+monkey during the grabbing. With $F = {dp\over dt}$ that means p is the same before and after (it's not constant, but there's no sudden change. But you have me in doubt now ... )