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Date Posted: 7/24/2008 12:28:56 PM  Status: Live
will rate high
Course Textbook Chapter Problem
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Question Details:
A rock of mass 0.3 kg is swung in a circular (vertical) path on a .25 meter string.  At the top of the path the angular speed is 12 rad/s.  What is the tension in the string at this point?
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Date Posted: 7/24/2008 1:03:20 PM  Status: Live
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Response:
Centripetal force required to keep the mass in a circular path
= m * ω2 * r = 0.3 * 122 * 0.25 = 10.8 N
 
As the rock is at the top of its path, some of the required centripetal force is supplied by the weight of the rock.
 
∴Weight of Rock = m g = 0.3 * 9.81 = 2.94 N
 
The tension in the string supplies the remaining centripetal force to continue the circular motion of the rock.
 
∴ Tension = 10.8 - 2.94 = 7.86 N



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