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Question: The radius of front wheel of arrangement shown in Fig. is a and that of rear wheel is b. If a dust p...

The radius of front wheel of arrangement shown in Fig. is a and that of rear wheel is b. If a dust particle driven from the highest point of rear wheel alights on the highest point of front wheel, the velocity of arrangement is

A

[g(b−a)(c−a+b)(c+a−b)]1/2\left\lbrack g\frac{(b - a)(c - a + b)}{(c + a - b)} \right\rbrack^{1/2}

B

g(c+a−b)4(b−a)g\frac{(c + a - b)}{4(b - a)}

C

[g(c+a−b)(c−a+b)4(b−a)]1/2\left\lbrack g\frac{(c + a - b)(c - a + b)}{4(b - a)} \right\rbrack^{1/2}

D

[g(c−a+b)4(c−a)]1/2\left\lbrack g\frac{(c - a + b)}{4(c - a)} \right\rbrack^{1/2}

Answer

[g(c+a−b)(c−a+b)4(b−a)]1/2\left\lbrack g\frac{(c + a - b)(c - a + b)}{4(b - a)} \right\rbrack^{1/2}

Explanation

Solution

Let t be time of flight of the particle from P to Q. Since e is the distance between the centres of the two wheels, the horizontal distance between the wheels.

ch = c2−(b−a)2\sqrt{c^{2} - (b - a)^{2}}

If v is the velocity of the carriage,

then ch = vt (horizontal range)

or vt = c2−(b−a)2\sqrt{c^{2} - (b - a)^{2}} or t = 1vc2−(b−a)2\frac{1}{v}\sqrt{c^{2} - (b - a)^{2}}

In this time, the particle has covered the vertical distance = 2(b - a).

∴ 2(b - a) = 12gt2\frac{1}{2}gt^{2}

or2(b−a)=12g[c2−(b−a)2]v22(b - a) = \frac{1}{2}g\frac{\left\lbrack c^{2} - (b - a)^{2} \right\rbrack}{v^{2}}or v2 = g4c2−(b−a)2(b−a)\frac{g}{4}\frac{c^{2} - (b - a)^{2}}{(b - a)}

or v = [g(c+a−b)(c−a+b)4(b−a)]1/2\left\lbrack g\frac{(c + a - b)(c - a + b)}{4(b - a)} \right\rbrack^{1/2}