A 290-g rock attached to a string travels 88.8 centimeters every 0.249 seconds on a horizontal circular path.

Question

A 290-g rock attached to a string travels 88.8 centimeters every 0.249 seconds on a horizontal circular path. The magnitude of the tension force in the string is 28.9 N. What is the diameter of this circular path in cm?

Answer 1

Hi Abdul,

This is circular motion problem. The equation for the centripetal force is: F=mv^2/r where

m = mass of the object

v = speed of the object

r = radius of the circle traversed

F = centripetal force (net force on the object)

The mass and force are both given, and we can calculate the speed from the fact that it travels 88.8 cm every 0.249 s. v = 0.888 m/0.249 s = 3.56 m/s.

Solving the equation for r we get:

r = mv^2/F = 0.29 kg*(3.56 m/s)^2 / 28.9 N

This yields a radius of 0.128 m. The problem asks for the diameter which is double that or 0.255 m or 25.5 cm.