STEP III 1990 question 13 solution

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STEP III 1990 question 13 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1990 question 13 solution

Consider a general point P which makes an angle $\theta$ with the upwards vertical.

The forces acting on the particle are its weight, acting downwards with magnitude mg, and the reaction force R acting towards O. Let the radius of sphere be r.

$R + mg\cos\theta = ma = \frac{mv^{2}}{r}$

$R = \frac{mv^{2}}{r} - mg\cos\theta$

The particle leaves the surface when R = 0.

$\frac{mv^{2}}{r} = mg\cos\theta \Leftrightarrow \frac{v^{2}}{r} = g\cos\theta$

$v^{2} = rg\cos\theta$

Since the velocity is tangential to the circle, the resolved velocity vector makes an angle of $\theta$ to the horizontal.

Therefore, the velocity and position vectors are:

$\textbf{v} = (v\cos\theta)\textbf{i} + (v\sin\theta - gt)\textbf{j}$

$\textbf{r} = (v\cos\theta t)\textbf{i} + t(v\sin\theta - \frac{1}{2}gt)\textbf{j}$

Let the collision occur at some point C.

We're trying to find the angle which $\theta$ has to exceed, so we can consider C at the same height as O, and find the resulting value of $\theta$ ; then $\theta$ has to be larger than this for the collision to be below O.

The horizontal distance to C is $r(1 + \sin\theta)$ ; the horizontal velocity is a constant $v\cos\theta$ . Therefore the collision occurs at $t = \frac{r(1 + \sin\theta)}{v\cos\theta}$ .

Inserting this value of t into the equation for the height:

$\displaystyle \textbf{r}_{j} = \frac{r(1 + \sin\theta)}{v\cos\theta} \left(v\sin\theta - \frac{rg(1 + \sin\theta)}{2v\cos\theta}\right)$

$\displaystyle = \frac{r(1 + \sin\theta)(2v^{2}\sin\theta\cos\theta - rg(1 + \sin\theta))}{2v^{2}\cos^{2}\theta}$

$\displaystyle = \frac{r(1 + \sin\theta)(2rg\sin\theta\cos^{2}\theta - rg(1 + \sin\theta))}{2rg\cos^{3}\theta} = \frac{r(1 + \sin\theta)(2\sin\theta\cos^{2}\theta - (1 + \sin\theta))}{2\cos^{3}\theta}$

The change in height is $-r\cos\theta$ . Equating the two you get:

$\displaystyle (1 + \sin\theta)(2\sin\theta(1 - \sin^{2}\theta) - 1 - \sin\theta) = -2( 1 - \sin^{2}\theta)^{2}$

When you multiply this all out and simplify, you end up with $2\sin^{3}\theta + 3\sin^{2}\theta - 1 = 0$ ; the only real solution to this is $\sin\theta = \frac{1}{2}$ , so when the collision is at the same height as O, $\theta = \frac{\pi}{6}$ .

For C to be below O, $\theta > \frac{\pi}{6}$ .

Alternate Method

Energy is conserved, so $\frac{1}{2}mv^{2} + mgr(1 + \cos\theta) = \frac{1}{2}mV^{2} + mgr$

Where v = velocity at P, V = velocity at C.

This simplifies to $V^{2} = v^{2} + 2gr(1 + \cos\theta) - 2gr = gr\cos\theta + 2gr\cos\theta = 3gr\cos\theta$ .

Using the velocity vector, we find $\displaystyle V^{2} = 3gr\cos\theta = (v\cos\theta)^{2} + \left(v\sin\theta - \frac{gr(1 + \sin\theta)}{v\cos\theta}\right)^{2}$

When this is simplified you end up with the same cubic equation in terms of $\sin\theta$ .

Solution by Dystopia.

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