## Friday, September 5, 2014

### Rotating projected ball

Inspired by a problem from brilliant.org A solid spherical ball with radius $R$ is projected from a rough ground with a velocity $v$ at angle $\theta$ with the horizontal, traveling to the right. At the same time it is given an angular velocity of $\omega$ clockwise such that the axis of rotation is perpendicular to the plane of projection.

At some point, the ball bounced on the ground, with coefficient of restitution $\frac{1}{2}$ and coefficient of friction is enough so that the ball did not slip while touching the ground. Find the initial angle such that the horizontal distance traveled by the ball to the point of second bounce be maximized.

### Solution

The initial velocity components are $v_x = v\cos\theta,v_y = v \sin \theta$. When the ball first touches the ground again, it will do so after a $t = 2v_y/g$ time, covering a distance of $d_1=tv_x = 2v_xv_y/g$.

At that point, the new velocity will have components $v'_x,v'_y$ respectively. We will calculate each component separately.

The vertical velocity is affected by the coeff. of restitution. Since the ground is assumed to be stationary (mass of the ball << mass of the Earth), then $v'_y = v_y / 2$

The horizontal velocity is affected by the friction. At the point of contact, the ball tends to go to the left, so the friction tends to go to the right. Suppose that the impulse caused by friction is $I$, then we have the following:

Linearly, the friction helps the linear motion: $$I = M(v'_x - v_x)$$ Rotationally, the torque caused by the friction slows down the angular motion: $$IR = \frac{2}{5}MR^2 (\omega - \frac{v'_x}{R})$$ (the last term is because we're assuming that the ball doesn't slip when it touches the ground). Solving for $v'_x$ we have: $$MR(v'_x - v_x) = \frac{2}{5}MR^2 (\omega - \frac{v'_x}{R})$$ $$v'_x - v_x = \frac{2}{5}R (\omega - \frac{v'_x}{R})$$ $$v'_x - v_x = \frac{2}{5}R\omega - \frac{2v'_x}{5}$$ $$v'_x = \frac{5}{7}( v_x+\frac{2}{5}R\omega ) = \frac{5v_x+2R\omega}{7}$$ Similar to the distance to the first bounce, the distance between first and second bounce is $$d_2 = 2v'_x v'_y/g = \frac{v_y(5v_x+2R\omega)}{7g}$$ Total distance is therefore: $$d_1+d_2 = \frac{2v_xv_y}{g} + \frac{v_y(5v_x+2R\omega)}{7g} = \frac{19v_xv_y + 2Rv_y\omega}{7g}$$ $$= \frac{v}{7g}(19v\sin\theta\cos\theta + 2R\sin\theta\omega)$$ Now we wish to find $\theta$ that maximizes: $$19v\sin\theta\cos\theta + 2R\omega\sin\theta$$ which can be done in various ways (calculus, double-angle formula etc).