Question

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When the centripetal forces of the two stones are equal, by substituting the value in the centripetal relation, the value of n can be calculated.

Answer 1

The Correct Option is C

Let v be the tangential speed of the heavier stone. Then, the centripetal force experienced by lighter stone is \((F_c)_{lighter}=\frac{m(nv)^2}{r}\) 
and that of heavier stone is \((F_c)_{heavier}=\frac{2mv^2}{(r/2)}\) 

But it is given that (F_c)_{lighter =} (F_c)_heavier

\(\therefore\) \(\frac{m(nv)^2}{r} = \frac{2mv^2}{(r/2)}\)

\(n^2\Bigg(\frac{mv^2}{r}\Bigg)=4\Bigg(\frac{mv^2}{r}\Bigg)\)

n² = 4

n = 2

Answer 2

Option C is correct. 

The centripetal force of both stones must be equal. 

From that, mass, velocity, and radius can be related and n is calculated by substituting the given relationship between the speed of stones in the centripetal force equation. 

\(\frac{m(nv)^2}{r} = \frac{2mv^2}{(r/2)}\)

nv₂ = 2v₂

n=2

Hence, the result is n = 2.

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is 2.