Answer
$\sqrt{\frac{Mrg}{m}}$
Work Step by Step
We can find the required speed as follows:
$T=Mg$
$T=\frac{mv^2}{r}$
Now $\frac{mv^2}{r}=Mg$
This can be rearranged as:
$v^2=\frac{Mgr}{m}$
$\implies v=\sqrt{\frac{Mrg}{m}}$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 186: 105
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