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.

### 2 Answers | Add Yours

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

**Sources:**

The ball is moving at 30 rpm.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes