A 0.4 kg object rotates in a vertical circle at the end of a 0.5 m string. What is the tension of the string at the bottom if the angular velocity there is 8.0 rad/s?
centripetal acceleration = R*w^2
weight = mg
R = radius of circle
w = angular velocity
The Attempt at a Solution
centripetal acceleration = (0.5 m)(8.0/s)^2 = 32 m/s^2
total (centripetal) force = (0.4 kg)(32 m/s^2) = 12.8 N (upwards)
weight = (0.4 kg)(9.8 m/s^2) = 3.9 N (downwards)
Upwards forces and accelerations will be considered positive, downwards ones negative.
total (centripetal) force = string tension - weight
12.8 N = string tension - 3.9 N
string tension = 12.8 N + 3.9 N = 16.7 N
In general terms:
string tension = mg + mRw^2
Unfortunately, the "official" solution is 13 N.
Am I wrong, or are my course materials wrong?
Will I be equally puzzled if I try to take the AP Physics B exam?