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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