Asked by Trevor
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?
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Answers
Answered by
Count Iblis
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.
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.
Answered by
Trevor
Is it vertical vs. Horizontal energy to get the Acceleration?
Answered by
Trevor
NVM you are relating GPE for Horizontal to KE of Horizontal.
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