How Is String Tension Calculated in Vertical Circular Motion?

[Kyriam]

Homework Statement

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?

Homework Equations

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?

 

[PhanthomJay]

Kyriam, welcome to PF! Your answer and method is good! I think you'll do OK in the exam.

 

[thomate1]

I cannot comment on the accuracy of your course materials or the AP Physics B exam. However, based on your calculations, your answer of 16.7 N seems to be correct. It is important to double check your calculations and make sure you are using the correct values for mass, radius, and angular velocity. It is also possible that there is a typo in the "official" solution. As a scientist, it is important to always question and double check information, so don't be afraid to ask your teacher or a peer for clarification.