Question

A ball of mass 0.25kg attached to the end of a string of length 1.96m is moving in a horizontal circle. The string will break if the tension is more than 25N. What is the maximum speed with which ball can be moved?
A.) 14 m/s
B.) 3 m/s
C.) 3.92 m/s
D.) 5 m/s

Answer 1

Hint: According to the question, the string will only break if the tension is greater than 25N which is the upward force that is binding the string and ball together. The outward force that is acting on the ball is the centripetal force. If the centripetal force is greater than the upward tension, only then the string will break that means we have to equate centripetal force and tension of string, to attain the maximum frequency that the ball can acquire.

Complete step-by-step answer:
We know that when an object moves in circular motion, centripetal force acts outwards, which is

$F = \dfrac{{m{v^2}}}{r}$

Where F is the centripetal force, m is the mass of the body revolving in circular motion, r is the radius of the circle formed by the object moving in circular motion and v is the velocity with which the object is moving.

Given that mass of the ball, $m = 0.25kg$
Tension of the string, $F = 25N$
Radius of the circle formed by ball, $r = 1.96m$
To calculate the maximum speed that the ball can acquire while rotating can be calculated by substituting all values in the above equation. We get,

$\Rightarrow F = \dfrac{{m{v^2}}}{r} \Rightarrow 25 = \dfrac{{0.25{v^2}}}{{1.96}} \Rightarrow {v^2} = \dfrac{{25(1.96)}}{{0.25}} \Rightarrow {v^2} = 196 \Rightarrow v = 14$

Here, there can be two values of velocities but we will take into consideration only the positive value because we only need the magnitude of the velocity.
Therefore, the maximum velocity that the ball can acquire is 14 m/s.
Hence, option (A) is the correct answer.

Note: Knowledge of centripetal force and the formula to compute it, is required. Maximum tension on the string shall be equated to the centripetal force to compute the maximum velocity with which the ball shall rotate.