STEP 1999 Paper 2 Question 11

By Neil Morrison (P1462) on Thursday, August 3, 2000 - 07:27 pm :

An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius

R

in a horizontal plane at a constant angular speed

ω

. A shell is fired from

O

, the centre of the circle, with initial speed

V

and angle of elevation

α

. Show that if

V^{2} < g R

, the no matter what the value of

α

, or what vertical plane the shell is fired in, the shell cannot hit the target.

[because, of course, the shell cannot travel $R$ , but I'm giving the whole question;]

Assume now that $V^{2} > g R$ and that the shell hits the target and let $β$ be the angle through which the target rotates between the time at which the shell is fired and the time of impact. Show that $β$ satisfies the equation:

$g^{2} β^{4} - 4 ω^{2} V^{2} β^{2} + 4 R^{2} ω^{4} = 0$

[I got this just by adding the bits up]

Deduce that there are exactly two possible values of $β$ .

[This is the bit I don't get. Why are there two values? The quadric equation would have to have its middle SP below the $β$ axis or have the other two SPs both sitting on the $β$ axis. And the former is not possible as the constant is < 0.]

Let $β_{1}$ and $β_{2}$ be the possible values of $β$ and let $P_{1}$ and $P_{2}$ be the corresponding points of impact. By considering the quantities $(β_{1}^{2} + β_{2}^{2})$ and $β_{1}^{2} β_{2}^{2}$ or otherwise show that the linear distance between $P_{1}$ and $P_{2}$ is $2 R \sin {[ω / g] (V^{2} - R g)^{1 / 2}}$

Neil M

By David Loeffler (P865) on Thursday, August 3, 2000 - 10:58 pm :

Well, I suppose that the quartic can be regarded as a quadratic in

β^{2}

. This will have either two, or zero, positive roots since the constant term is positive. In fact, it has exactly two, because

V^{2} > g R

, so the minimum is less than the minimum of

g^{2} β^{4} - 4 g β^{2} r ω^{2} + 4 r^{2} ω^{4} = (g β^{2} - 2 r ω^{2})^{2} = 0

Thus there are two distinct positive values of $β^{2}$ , giving rise to two distinct positive (and two negative) roots of the quartic.

Hope this helps.

David

By Neil Morrison (P1462) on Friday, August 4, 2000 - 10:18 am :

I think I had got to that stage, but I was trying to win the logic battle. But the problem is it says exactly two possible values. And in your solution, we have 2 positive and 2 negative, the negative ones (say

A

and

B

) can be regarded as

2 π + A

and

2 π + B

. The only way I could resolve this is to show that the "

2 π +

" values end up being equal to the original positive values, so in effect there are only two distinct solutions.

Thanks anyway

Neil M

By David Loeffler (P865) on Friday, August 4, 2000 - 01:55 pm :

That would actually be impossible, surely, since

π

is transcendental? But the physical situation implies that we're not looking for "position" angles which are mod

2 π

, but for the angle the target has actually moved through. This will not be mod

2 π

as, for example,

3 π

would correspond to one and a half circuits, not to half a circuit. The target won't turn around and go backwards, so perhaps that is why we can discard the negative solutions.

By Neil Morrison (P1462) on Friday, August 4, 2000 - 04:43 pm :

But the net position of the target will be the same after

3 π

as if it had moved

π

backwards.

However, I agree that this will not work. It was the only way I could think of to justify the question. I think its badly worded.

Neil M