Question

A 0.40-kg mass, attached to the end of a 0.75-m string, is whirled around in a circular horizontal path. If the maximum tension that the string can withstand is 450 N, then what maximum speed can the mass have if the string is not to break?

Answer 1

To solve this problem we will apply the concepts given from the circular movement of the bodies for which we have that the centripetal Force is defined as a product between the mass and the velocity squared at the rate of rotation, mathematically this is

$F_c = \frac{mv^2}{r}$

Where,

m = Mass

v = Velocity

r = Radius

Our values are given as

$m = 0.4kg\\r = 0.75m\\F_c = 450N$

Rearranging to find the velocity we have that,

$F_c = \frac{mv^2}{r}$

$v = \sqrt{\frac{F_c * r}{m}}$

$v = \sqrt{\frac{450 * 0.75}{0.4}}$

$v = 29.0474m/s$

Therefore the  maximum speed can the mass have if the string is not to break is 29m/s