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# A ball of mass 0.25kg attached to the ends of a string of length 1.96m is rotating in a horizontal circle. The string will break, if tension is more than $25N$. What is the maximum velocity with which the ball can be rotated?

Last updated date: 15th Jun 2024
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Hint: Centrifugal Force acts on every object moving in a circular path when viewed from a rotating frame of reference. The centrifugal force also depends on the mass of the object, the distance from the centre of the circle and also the speed of rotation. If the object has more mass, the force of the movement and the speed of the object will be greater. If the distance is far from the centre of the circle the force of the movement will be more.

Formula Used:
The mathematical formula for Centrifugal force is given as the negative product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:
${F_c} = \dfrac{{ - m{v^2}}}{r}$
In this mathematical formula, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius.

In this above numerical problem, the mass of the ball is given equal to $0.25kg$ and the force of tension is given equal to $25N$. Now, the radius of the string is given equal to $1.96m$.
${v^2} = \dfrac{{F \times r}}{m}$
${v^2} = \dfrac{{25 \times 1.96}}{{0.25}} = 196$
$v = 14m{s^{ - 1}}$