Solution to 1986 Problem 37

The centripetal acceleration must be
\begin{align*}a = \frac{v^2}{r} = \omega^2 r\end{align*}
So, m \omega^2 r must be the horizontal component of the tension in the cord. The mass is not accelerating in the vertical direction, so the vertical component of the tension must be mg. So, the tension in the cord is:
\begin{align*}F_T = \sqrt{(m g)^2 + (m \omega^2 r)^2 } = \boxed{m \sqrt{g^2 + \omega^4 r^2}}\end{align*}
Therefore, answer (E) is correct.

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