A small block of mass m is tied to a string of length L and whirled in a circle on a smooth horizontal table. The string will break if the tension exceeds a certain maximum value. The maximum angular velocity with which the block can be whirled without breaking the string is ₀ . If the string is shortened to a length of L 2 , what will be the new maximum angular velocity before the string breaks?

Step-by-step solution

The tension in the string provides the necessary centripetal force for the circular motion of the block. The maximum tension T_ max the string can withstand before breaking is given by: T_ max = mL ₀^2 When the length of the string is halved to L 2 , the maximum tension the string can bear remains the same. Let the new maximum angular velocity be ' . T_ max = m ( L 2 ) ( ')^2 Equating the two expressions for T_ max : mL ₀^2 = m ( L 2 ) ( ')^2 ₀^2 = ( ')^2 2 ( ')^2 = 2 ₀^2 ' = 2 ₀ . Answer: 2 ₀