A light string can
support a stationary hanging load
of 25.0 kg before
breaking. An object
of mass m 5 3.00 kg
attached to the string
rotates on a frictionless, horizontal table
in a circle of radius
r 5 0.800 m, and
the other end of the
string is held fixed
as in Figure P6.1. What range of speeds can the object
have before the string breaks?
The tension in the string is responsible for keeping the object in circular motion. At the maximum speed, the tension in the string will be equal to its breaking strength. We can use centripetal force to find the maximum speed of the object:
F_c = m v^2 / r
where F_c is the centripetal force, m is the mass of the object, v is its velocity, and r is the radius of the circle.
The tension in the string is equal to the centripetal force:
T = F_c = m v^2 / r
When the tension is equal to the breaking strength of the string, the string will break:
T = 25.0 kg * 9.8 m/s^2 = 245 N
Solving for v:
v^2 = T * r / m
v^2 = 245 N * 0.800 m / 3.00 kg
v^2 = 65.3 m^2/s^2
v = sqrt(65.3) m/s
v = 8.09 m/s
Therefore, the object can have a speed of up to 8.09 m/s before the string breaks.