Question 177: Work, Energy and Power & Circular Motion

Question

Two stones of masses m and 3 m are whirled in horizontal circles, the heavier one in a radius $\left(\frac{\mathrm{r}}{3}\right)$ and lighter one in a radius r . The tangential speed of lighter stone is ' $n$ ' times the value of heavier stone. When the magnitude of centripetal force becomes equal the value of $n$ is

Accepted Answer

We have
two stones
:
Stone A (lighter): mass = $m$, radius of circular motion = $r$, tangential speed = $v_A$.
Stone B (heavier): mass = $3m$, radius of circular motion = $	frac{r}{3}$, tangential speed = $v_B$.
It is given:
$ v_A = n\,v_B $
We need: condition when centripetal forces are equal in magnitude.
Step 1. Write centripetal force formula
$ F = \frac{mv^2}{R} $
Step 2. Forces for two stones
For Stone A:
$ F_A = \frac{m v_A^2}{r} $
For Stone B:
$ F_B = \frac{(3m) v_B^2}{	frac{r}{3}} = \frac{3m v_B^2}{(r/3)} = \frac{9m v_B^2}{r} $
Step 3. Equating forces
$ F_A = F_B $
$ \frac{m v_A^2}{r} = \frac{9m v_B^2}{r} $
Cancel $m$ and $r$:
$ v_A^2 = 9 v_B^2 $
$ \frac{v_A}{v_B} = 3 $
So:
$ v_A = 3 v_B $
That means $n = 3$.
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Answer: 3 (Option B)

Answer

4

Answer

3

Answer

2

Answer

1

Work, Energy and Power & Circular Motion

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