Question

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius $\frac{r}{2}$ 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) 3(b) 4(c) 1(d) 2

Answer 1

If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: ... T is the tension in the string. ... Here length of string 'l ' is connected to a particle of mass m . after moving of string in a circle with speed v.

Answer 2

Answer:

The value of n is 2.

(d) is correct option.

Explanation:

Given that,

Mass of first stone = m

Mass of second stone = 2m

Radius of lighter stone = r

Radius of heavier $r= \dfrac{r}{2}$

Speed of lighter stone = nv

We need to calculate the value of n

Using centripetal force

$f_{h}=f_{l}$

$\dfrac{mv^2}{r}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{2m\times v^2}{\dfrac{r}{2}}=\dfrac{m(nv)^2}{r}$

$\dfrac{4}{r}=\dfrac{n^2}{r}$

$n =2$

Hence, The value of n is 2.