## Monday, September 29, 2014

### Projectile fired

A projectile fired from the ground passes over two points $A$ and $B$ that are horizontally $d$ apart. Point $A$ is $2h$ high from the ground whereas point $B$ is horizontally $3h$ high from the ground. What's the minimum speed of projection?

## Friday, September 19, 2014

### Inverse distances to points

Let $P$ and $Q$ be two distinct points on a plane. For any given point $X \neq P,Q$, define two functions $f(X)$ and $g(X)$ as follows: $$f(X) = \frac{1}{PX} + \frac{2}{QX}$$ $$g(X) = \frac{2}{PX} + \frac{1}{QX}$$ Now for any positive number $t$, let $S(t)$ be the set of all points $Y$ such that $f(Y) \leq tg(Y)$.

Prove that $S(t)$ is bounded

Prove that $S(t)$ is convex. That is, if $A,B$ are in $S(t)$ then $AB$ is in $S(t)$.

Identify all points $X$ on $S(t)$ such that $PX+QX$ is minimum.

Prove that the area of $S(t)$ is a convex function of $t$.

### Powered distances

Let $p$ and $q$ be two infinitely long non-parallel lines on a plane. For any point $X$, let $d_p(X), d_q(X)$ denote distances from $X$ to $p$ and $q$ respectively. For any positive number $r$, let $S(r)$ be the set of all points $Y$ such that $d_p(Y)^{2015} + d_q(Y)^{2015} \leq r$.

Prove that $S(r)$ is bounded.

Prove that $S(r)$ is convex. That is, if $A$ and $B$ are in $S(r)$ then $AB$ is in $S(r)$.

Identify all points in $S$ such that $d_p(X)^{2014} + d_q(X)^{2014}$ is maximum.

Prove that the area of $S(r)$ is a convex function of $r$.

## Friday, September 12, 2014

### Ball hitting an inclined plane

A ball with mass $m$ and radius $R$ is rolling without slip with velocity $v$ on a horizontal plane. It then hits an inclined plane with angle $\theta$, after which it continues rolling without slip upwards along the plane. How far does the ball travel along the plane before it stops?

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.

## 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).

## Thursday, September 4, 2014

### Pencil rolling down the plane

Inspired by IPhO 1998.

A pencil that's made of homogeneous wood is shaped like a regular hexagonal prism, with hexagon side $s$ and mass $M$. The pencil rolls down an inclined plane. Find the moment of inertia.

Assume that the sides are slightly concave so that only the vertices of the hexagons are touching the plane at any given moment.

### Solution

First we find the moment of inertia of a equilateral prism about its center. Suppose the triangle side is $s$. By dimensional analysis, we know that the moment inertia $I$ must have the form $I = kMs^2$ for some $k$. Also note that the equilateral triangle is a superposition of 4 smaller triangles.

The moment of inertia from the triangle in the center is $k.M/4.(s/2)^2 = kMs^2 / 16$.

The moment of inertia from each of the outer smaller triangle can be found using parallel axis theorem. From geometry, we know the displacement of the axes is $s / \sqrt{3}$. So: $$I_t = \frac{kMs^2}{16} + \frac{M}{4} . (\frac{s}{2\sqrt{3}})^2 = \frac{Ms^2}{16} (k + \frac{1}{3})$$ Therefore our original moment of inertia should be equal to the superposition of four triangles: $$kMs^2 = \frac{kMs^2}{16} + 3\frac{Ms^2}{16} (k + \frac{1}{3})$$ $$k = \frac{k}{16} + \frac{3k+1}{16}$$ $$k = \frac{1}{12}$$ So the moment of inertia of an equilateral prism about its tip is $$\frac{Ms^2}{12} + M(\frac{s}{\sqrt{3}})^2 = \frac{5Ms^2}{12}$$ Thus the moment of inertia for six of these triangles about their tip (i.e. hexagon about the center) is $6 . 5/12 . M/6 s^2 = 5Ms^2/12$ So the moment of inertia of the hexagon about the edge is $$\frac{5Ms^2}{12} + Ms^2 = \frac{17}{12}Ms^2$$

### Second Solution

Alternatively we can find the moment of inertia of a triangular prism about its tip by dividing it into a series of slabs, where each slab does not touch the rotational axis. Each slab is infinitesimally thin (having thickness of $dx$) where $x$ is the distance from the axis to the slab. The slab would have a dimension of $l \times dx \times s$ where $x = \sqrt{3}s / 2$ so $s = 2x / \sqrt{3}$.

Suppose the density of the wood is $\rho$. The mass of the slab is then $m=\rho.l.s.dx$. The m.o.i of the slab around the center is $$\frac{m}{12}(dx^2 + s^2) \simeq \frac{ms^2}{12}$$ The m.o.i of the slab around the axis that's $x$ away is then $$\frac{ms^2}{12} + mx^2 = \rho.l.s.dx.(\frac{s^2}{12} +x^2) = \rho.l.\frac{20}{9\sqrt{3}} x^3 dx$$ The triangle is a superposition of these slabs, with $x$ ranging from $0$ to $\frac{\sqrt{3}}{2}S$ $$I = \int_0^\frac{\sqrt{3}S}{2} \rho.l.\frac{20}{9\sqrt{3}} x^3 dx = \rho.l.\frac{20}{9\sqrt{3}} . \frac{9/16 S^4}{4} = \frac{5}{16\sqrt{3}} \rho.l.S^4$$ But then $\rho = M / (l.S^2.\sqrt{3}/4) = 4M / (\sqrt{3}lS^2)$ so: $$I = \frac{5}{16\sqrt{3}} l.S^4 . \frac{4M}{\sqrt{3}lS^2} = \frac{5MS^2}{12}$$ The moment of inertia for the hexagon about the edge follows easily just like in the first solution.