Question 11.12: A circus trick In a circus trick, a performer of mass m caus...
A circus trick
In a circus trick, a performer of mass m causes a large ball of mass M and radius a to accelerate to the right (see Figure 11.12) by running to the left on the upper surface of the ball. The man does not fall off the ball because he maintains this motion in such a way that the angle α shown remains constant. Find the conditions neccessary for such a motion to take place.
Suppose the motion is planar and that, at time t, the ball has velocity V in the i-direction and angular velocity ω (= V/a) around the axis {O, j}. If the man is to maintain his position on the ball, then he must run up the surface of the ball (towards the highest point) with velocity V. This maintains his vertical height and his acceleraton is then the same as that of the ball, namely (dV/dt)i.
The equations of motion for the man are therefore
\begin{aligned}m \frac{d V}{d t} &=N_{2} \sin \alpha-F_{2} \cos \alpha \\0 &=N_{2} \cos \alpha+F_{2} \sin \alpha-m g\end{aligned}
and the equations of motion for the ball are
\begin{aligned}M \frac{d V}{d t} &=F_{1}-N_{2} \sin \alpha+F_{2} \cos \alpha \\0 &=N_{1}-N_{2} \cos \alpha-F_{2} \sin \alpha-M g \\I_{O} \frac{d \omega}{d t} &=a F_{2}-a F_{1}\end{aligned}
where I_{O} is the moment of inertia of the ball about {O, j}.
These five equations, together with the rolling condition V = ωa, are sufficient to determine the six unknowns d V / d t, d \omega / d t, F_{1}, N_{1}, F_{2} \text { and } N_{2}. After some algebra, the solution for the forward acceleration of the ball turns out to be
\frac{d V}{d t}=\frac{m g \sin \alpha}{M+\left(I_{O} / a^{2}\right)+m(1+\cos \alpha)}.
Hence the motion is possible for any acute angle α provided that the performer can accelerate relative to the ball with this acceleration. For the case in which the ball is hollow, the masses of the man and the ball are equal, and α = 45°, the acceleration required is approximately 0.21 g.