Question
Question: A stone of mass 22kg is tied to a string of length 0.5m. If the breaking tension of the string is 90...
A stone of mass 22kg is tied to a string of length 0.5m. If the breaking tension of the string is 900N then the maximum angular velocity the stone can have in a uniform circular motion is
A) 30 rad/s
B) 20rad/s
C) 10rad/s
D) 25rad/s
Solution
Imagine you are having a small stone attached to a string and you are rotating the stone. When the stone is rotating it is experiencing a force which is directed towards the centre i.e. towards you (since you are rotating the stone). The force in the tension of the string will be equal to the force applied by you in rotating the stone (that’s why the stone is able to rotate).
Complete step by step answer:
The formula for centripetal force is:
Fc=rmv2
Write the above formula in terms of angular velocity.
Fc=rmv2=mrω2;
Now, Put in the given values for mass, length of the rope and the force due to the tension in the string. As the stone is being pulled at the centre the acceleration here would be equal to the centripetal acceleration and since the above analogy is true the tensional force would be equal to the centripetal force.
⇒Fc=mrω2;
On substituting the corresponding values,
⇒900=22×0.5×ω2;
keep the angular velocity ω2 on the RHS and take the rest to LHS:
⇒900=11×ω2;
Divide the numerator with the denominator and solve:
⇒11900=ω2;
To remove the square on the RHS, take the square root of the value given in the LHS.
⇒ω=81.81;
We get the value as:
⇒ω=9.0444≃10rad/s;
Hence, Option (C) is correct. The maximum angular velocity the stone can have in a uniform circular motion is 10 rad/s.
Note:
Here the stone is undergoing centripetal motion, so, apply the formula for the centripetal force which is linear in nature, and then convert it in terms of angular velocity where the linear velocity and angular velocity is related by “ω2=r2v2”. Put the given value and find the unknown.