Posted by **Trevor** on Sunday, November 18, 2007 at 5:55pm.

An object is swinging from a point that is horizontal compared to the lowest point the object will reach (where the rope is vertical). If the length of the rope is 2 meters and the mass of the object is 30 kg, what is the tension in the rope at the bottom of the swing?

Need Step by step solution!

- Physics -
**Count Iblis**, Sunday, November 18, 2007 at 6:10pm
Drop in potential energy from horizontal to vertical orientation is

2 meters *m* g

This must equal 1/2 m v^2, because velocity is zero when the rope is horizontal. This means that:

v^2 = 4 meters*g --->

v^2/(2meters) = 2 g is the centripetal acceleration.

Newton's second law:

F = m a

applied to the object gives:

-mg + T = 2mg

where T is the tension and we take the positive direction to be upward.

The tension is thus T = 3 m g.

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