Question

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answer 1

When the mass which is attached to the string is rotating in circular motion, it will be acted upon by the centripetal force. If we observe from the frame of stone, centrifugal force will be acting. Now that centripetal force and centrifugal force is the same in magnitude. If we consider centripetal force, it will be provided by tension in string. So by equating tension to centripetal force we will solve this problem.

Formula used:
\eqalign{ & {F_C} = \dfrac{{m{v^2}}}{r} \cr & v = r\omega \cr & {F_C} = mr{\omega ^2} \cr}

Complete step-by-step answer:
When an object is rotating in a circle its direction of velocity keeps on changing. When velocity changes there will be an acceleration and this contributes for the force. Now the force which is responsible for the change in direction in case of circular motion is called a centripetal force.
Centripetal force is given by the formula ${F_C} = \dfrac{{m{v^2}}}{r}$
Where ‘v’ is the magnitude of velocity of rotation and ‘r’ is the radius of the circle of the circular motion and ‘m’ is the mass of the object which is rotating in a circle.
We can express the linear velocity ‘v’ in terms of angular velocity ‘$\omega$’ as $v = r\omega$
Then the formula for centripetal force will become ${F_C} = mr{\omega ^2}$
In this question $\omega = 40{\text{rev}}/\min = \dfrac{{40 \times 2\pi }}{{60}} = 4.2rad/s$
In case of rotation of stone the centripetal force is provided by the string and hence centripetal force will be the tension in the string. Let tension in the string be T
We have

\eqalign{ & T = mr{\omega ^2} \cr & \Rightarrow T = 0.25 \times 1 \times {(4.2)^2} = 4.4N \cr}
So tension in the first case will be 4.4 Newton
In the second case tension is given as 200N and we are asked to find out the velocity of rotation
${F_C} = \dfrac{{m{v^2}}}{r}$
\eqalign{ & \Rightarrow T = \dfrac{{m{v^2}}}{r} \cr & \Rightarrow T = \dfrac{{0.25 \times {v^2}}}{1} \cr & \Rightarrow 200 = 0.25 \times {v^2} \cr & \Rightarrow 800 = {v^2} \cr & \Rightarrow 28.28m/s = v \cr & \Rightarrow v = 28.28m/s \cr}
So in the second case the velocity will be 28.28 meters per second

Note: The question given is the case of constant magnitude of velocity and hence centripetal force will be constant. There are some cases where magnitude of velocity depends on the time i.e speed depends on the time and it varies with time. Then centripetal force will be varying at every instant and we come across a new term called tangential acceleration.