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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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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted May 19, 2012 at 7:04 PM via web

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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 30 rpm. If the spring was 55 cm long initially what is the spring constant.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 19, 2012 at 7:20 PM (Answer #1)

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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

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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted May 19, 2012 at 7:07 PM (Answer #2)

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The ball is moving at 30 rpm.

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