# Acrobat grabbing monkey

## Homework Statement

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?

center of mass

## The Attempt at a Solution

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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?

BvU
Homework Helper
When grabbing, y1(te) = h too.

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

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:
BvU
Homework Helper
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!

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## ?

BvU