Question

A string 0.1m cannot bear tension more than 100 N, it is tied to a body of mass 100g and rotated in a horizontal circle. The maximum angular velocity can be.

Answer 1

Hint: Whenever anybody is whirled with a thread or string and moved in a circular motion then tension depends upon the \[\omega \] (angular velocity) of the bod which is balanced by centripetal force. As the angular velocity increases the tension in the string also increases. At the same time tension also depends on the mass of the tied body with string.

Complete step by step answer:
In the given situation the tension in the string will be equal to centripetal force.
Formula for centripetal force:
\[{F_c} = \dfrac{{m{v^2}}}{r} = m{\omega ^2}r\]
Where:
\[{F_c} = \] is the centripetal force
\[m = \] is the mass of the body
\[\omega = \] is the angular velocity
\[r = \] length of string
So according to question:
\[T = m{\omega ^2}r\]
Putting given values
\[100 = 0.1 \times {\omega ^2} \times 0.1\]
\[{\omega ^2} = \dfrac{{100}}{{0.1 \times 0.1}}\]
\[{\omega ^2} = 10000\]
\[\omega = 100rad/s\]
Thus, angular velocity of the string should be less than 100rad/s.

Note: Here if the mass was rotated at some angle with the vertical then only a particular component of tension would provide centripetal force and rest would balance the vertical motion force(gravity), however in this case it is a pure horizontal circular motion.