Also determine the maximum speed $v$ such that the ball does not lose contact with the plane (rebound).

### Solution

At the point of contact, there is a change of pivot point. Suppose that the speed after the impact is $v'$ then the angular speed is $v'/R$. Immediately before impact, the ball's center of mass travels horizontally, while immediately after, it travels parallel to the plane. Thus there is an impulse. Suppose we denote $F_x, F_y$ to be the horizontal and vertical component of the impulse respectively. The direction of $F_y$ is upwards, while $F_x$ is against the initial motion of the ball. The impulse occurs at the first point of contact with the inclined plane, because this is where the normal force from the plane acts on the ball. The impulse from the gravity is negligible because we assume instantaneous / momentary contact time.Change in vertical linear momentum: $$I_y = mv'\sin \theta$$ Change in horizontal linear momentum: $$I_x = m(v - v'\cos \theta)$$ Change in angular momentum. Remember that $I_y$ goes against the new rotational motion while $I_x$ helps it. $$I_x R \cos \theta - I_y R \sin \theta = \frac{2}{5}mR^2 (\frac{v'}{R} - \frac{v}{R})$$ We have three equations and three unknowns: $I_x, I_y, v'$. Solving for $v'$: $$I_x \cos \theta - I_y \sin \theta = \frac{2}{5}m(v'-v)$$ $$(v - v'\cos \theta) \cos \theta - v' \sin^2 \theta = \frac{2}{5}(v'-v)$$ $$v' = \frac{2+5\cos\theta}{7} v$$ Sanity check, if $\theta = 0$ then $v' = v$, and if $v'$ decreases with $\theta$.

From here we can use the law of conservation of energy, since post-collision there's no energy lost to friction. $$mgh = \frac{1}{2}mv'^2 + \frac{1}{2} \frac{2}{5}mR^2 \frac{v'^2}{R^2}$$ $$gh = \frac{7}{10} (\frac{2+5\cos\theta}{7} v)^2 = \frac{10(2+5\cos \theta)^2}{7} v^2$$ $$h = \sqrt{\frac{10}{7g}} (2+5\cos \theta)v$$ The distance traveled along the plane is $d = h / \sin \theta$ $$d = \sqrt{\frac{10}{7g}} (2+5\cos \theta) \frac{v}{\sin \theta}$$ The instant after the impact, the ball rotates around the new point of contact. The combination of weight and normal force supplies the necessary centripetal force. We express everything in the direction perpendicular to the plane: $$mg \cos \theta - N = \frac{mv'^2}{R}$$ $$N = m(g \cos \theta - \frac{v'^2}{R}$$ For $N \geq 0$ we must have that: $$g \cos \theta \geq \frac{(2+5\cos\theta)^2}{49R} v^2$$ $$v \leq \frac{7 \sqrt{gR\cos \theta}}{2+5\cos\theta}$$

### Alternative Solution

At the point of contact, there is a change of pivot point. Suppose that the speed after the impact is $v'$ then the angular speed is $v'/R$. Immediately before impact, the ball's center of mass travels horizontally, while immediately after, it travels parallel to the plane. Let $X$ be the new pivot point. We consider the angular momentum about $X$ before and after the impact. The force exerted by plane on the ball is at $X$ so there's no torque around $X$, thus these two angular momenta must be equal.Note that the m.o.i of a ball rotating around it's edge is $\frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.

The angular momentum before the impact depends on two things: the angular momentum of the center of mass (c.o.m) and the angular momentum of the ball around the c.o.m. The c.o.m was traveling horizontally, forming angle $\pi/2 - \theta$ with the radius vector. The $r \times v$ term is $vR \cos \theta$, thus the angular momentum of the c.o.m is $mvR \cos \theta$.

The angular momentum of the ball around the c.o.m is $\frac{2}{5}mR^2 v/R = \frac{2}{5}mvR$. These two momenta have the same direction so we can simply add them: $$ L= mvR (\cos \theta + \frac{2}{5})$$ Now the angular momentum after the impact is a simple rotation on the edge with the new speed: $$ L' = \frac{7}{5}mR^2 \frac{v'}{R} = mv'R \frac{7}{5} $$ So we have: $$ (\cos \theta + \frac{2}{5}) v = \frac{7}{5} v'$$ $$v' = \frac{2+5\cos\theta}{7} v$$ The rest is same as the first solution.

## No comments:

## Post a Comment