Question

Two stones of masses $m$ and $2m$ are whirled in horizontal circles, the heavier one in a radius, $\dfrac{r}{2}$ 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. Find the value of $n$?
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint: In the question, the relation between the mass, radius and speed of the two stones are given. When the centripetal forces of the two stones are equal in the given condition, by substituting the given relation in the centripetal relation, the value of $n$ can be calculated.

Useful formula:
The centripetal force on the object revolving with the radius,
$C.F = \dfrac{{m{v^2}}}{r}$
Where, $C.F$ is the centripetal force of the object, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius in which the object rotates.

Given data:
The mass of the first stone is $m$
The mass of the second stone is $2m$
The radius of the rotation of the first stone is $r$
The radius of the rotation of the second stone is $\dfrac{r}{2}$
The speed of the first stone is ${v_1}$
The speed of the second stone is ${v_2}$

Complete step by step solution:
From the question, it is clear that the two stones have the same centripetal force.
Thus,
$C.{F_1} = C.{F_2}\;....................................\left( 1 \right)$
Where, $C.{F_1}$ is the centripetal force of the first stone and $C.{F_2}$ is the centripetal force of the second stone.

The centripetal force on first stone,
$C.{F_1} = \dfrac{{m{v_1}^2}}{r}$

The centripetal force on second stone,
$C.{F_2} = \dfrac{{2m{v_2}^2}}{{\left( {\dfrac{r}{2}} \right)}}$
Substitute the values in equation (1), we get
$\dfrac{{m{v_1}^2}}{r} = \dfrac{{2m{v_2}^2}}{{\left( {\dfrac{r}{2}} \right)}}\;..............................................\left( 2 \right)$
It is also given in the question; the speed of the first stone is $n$ times the speed of the second stone.
Thus,
${v_1} = n{v_2}$
Substitute the above relation in equation (2), we get
$\dfrac{{m{{\left( {n{v_2}} \right)}^2}}}{r} = \dfrac{{2m{v_2}^2}}{{\left( {\dfrac{r}{2}} \right)}}\; \\\ \dfrac{{m{{\left( {n{v_2}} \right)}^2}}}{r} = \dfrac{{4m{v_2}^2}}{r}\; \\\ {\left( {n{v_1}} \right)^2} = 4{v_2}^2 \\\$
Taking square root on both sides, we get
$\sqrt {{{\left( {n{v_2}} \right)}^2}} = \sqrt {4{v_2}^2} \\\ n{v_2} = 2{v_2} \\\ n = 2 \\\$

Hence, the option (B) is correct.

Note: The stone with the higher mass tends to rotate in less radius compared to the stone of lower mass. That makes the velocity of the lighter stone is more than the heavier one. Since the mass, the radius parameter between the two stones are doubled and half respectively. The velocity value of lighter mass gets doubled than heavier one.