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A ball of 2.3 kg attached to a spring is moving in a circular path of radius 60 cm at...
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The ball of 2.3 kg is attached to a spring and it is moving in a circular path of radius 60 cm at 30 rpm. The initial length of the spring is 55 cm.
The angular velocity of the ball is w = (30*2*pi)/60 rad/s
w = pi rad/s
As the ball moves it experiences a force of m*v^2/r = m*w^2*r.
Substituting the mass and angular velocity F = 2.3*pi^2*0.6 = 13.62 N. As the ball is attached to a spring that had an initial length of 55 cm but is now 60 cm long, the force of 13.62 N is equal to k*x = 0.05*k
k*0.05 = 13.62
=> k = 272.4 N/m
The spring constant of the spring is 272.4 N/m
Posted by justaguide on May 19, 2012 at 7:20 PM (Answer #1)
Valedictorian, Dean's List
The ball is moving at 30 rpm.
Posted by lxsptter on May 19, 2012 at 7:07 PM (Answer #2)
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