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Problem 61:  Kinematics of rotating target ( ) 1999 Paper II

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 this circle, with initial speed V and angle of elevation α.

Show that if V 2 < gR, then no matter what the value of α, or what vertical plane the shell is fired in, the shell cannot hit the target.
Assume now that V 2 > gR and that the shell hits the target, and let β be the (positive) 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 g2β4 4ω2V 2β2 + 4R2ω4 = 0.

Deduce that there are exactly two possible values of β.

Let β1 and β2 be the possible values of β and let P1 and P2 be the corresponding points of impact. By considering the quantities (β12 + β22) and β12β22, or otherwise, show that the linear distance between P1 and P2 is
2Rsinω gV 2 Rg.


The rotation of the target is irrelevant for the first part, which contravenes the setters’ rule of not introducing information before it is required. In this case, it seemed better to describe the set-up immediately — especially as you are asked for a familiar result.

Remember, when you are considering the roots of the quartic (which is really a quadratic in β2), that the question gives β > 0.

The hint in the last paragraph (‘by considering ...’) is supposed to direct you towards the relation between the coefficients in a quadratic equation and the sum and product of the roots; otherwise, you get into some pretty heavy algebra.

Solution to problem 61

(i) This part is about the range of the gun. If the shell lands at distance x, then

x = (V cosα)t,0 = (V sinα)t 1 2gt2.

Eliminating t gives xg = V 2 sin2α. The range r is the largest value of x, which given by rg = V 2. The shell cannot reach the target (for any angle of elevation) if R > r, so the shell cannot hit its target if V 2 < Rg.

(ii) The time of flight is R(V cosα), and this must also equal the time for the target to rotate through β, i.e. βω. Thus cosα = Rω(βV ). Substituting this into the range equation Rg = 2V 2 sinαcosα gives

Rg = 2V 2Rω βV 1 R2ω2 β2V 2 1 2 .

Squaring both sides of the equation, and simplifying, leads to the given quadratic in β2. Solving the quadratic using the quadratic formula gives

g2β2 = 2ω2 V 2 ±V 4 g2 R2 .

Since V 2 > gR, both the values of β2 roots are real and positive, but they lead to only two relevant values of β since we only want positive values.

(iii) The linear distance between the two points of impact is 2Rsin 1 2(β2 β1). Now β12 + β22 = 4ω2V 2g2 and β12β22 = 4R2ω4g2 (sum and product of the roots of the quadratic), so

(β2 β1)2 = β 22 + β 12 2β 1β2 = 4ω2V 2 g2 4Rω2 g ,

which leads to the given result.


Why do we get two values for β? In order to hit the large circle, if it is in range, we can fix any speed of projection and choose the angle of projection appropriately. In general, there are two possible angles of projection, one above 45 and one below 45 (satisfying sin2α = RgV 2).

However, in this question, we want to hit the circle essentially at a given time (when the target is at the landing place), which means that V cosα is fixed, though of course it is a little more complicated because the position of the target is fixed but unknown. But still, the two values of β, corresponding to two different angles of elevation, emerge in the same way.

Did you notice the slight inaccuracy in the question? The final ‘distance’ could be negative (if for example πg < ωV 2 Rg < 2πg). Mod signs are needed.