Question

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.1-m-long rigid, massless rod. The rod is rotating cw about its geometric center at 17 rpm. What force should be applied to bring this structure to a stop in 4.4 seconds, if the force can only be applied at a location 0.25 m from the geometric center?

Answer 1

Since in the question it is mentioned that rotation is about geometrical center of rod,

Assume 1kg ball is at x=0 position and 2kg ball is at x=1.1m position.

So, rotation will be around X=1.1/2 m=0.55 m

since rotation is with 17 rotation per minute,(assuming clockwise rotation is positive)

angular velocity = 17 *2*pi rad/minute

= 1.779 rad/s

now to stop the rotation in 4.4 s, angular velocity will be 0 after 4.4 sec

so, angular acceleration, can be calculated by

so

torue required,

and

I =m1*(distance from geometrical center)^2 + m2*(distance fromgeometrical center)^2

since the force is acting 0.25m from geometrical center, And also to cancel net force on the rod you need to apply force on both side from geometrical center. so, that both force can cancel. So now F will produce torque. So, force applied at the both end

So, 0.734 N force is applied both side 0.25m from geometrical center of rod such that both force will produce same direction of torque.

good luck

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