# An object is rotating at 30 rpm attached to a string that is 10 m long. What is the work required to bring the object 5 m closer while it rotates at the same rate.

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### 1 Answer

An inertia of an object that is rotating in a circular path tends to make it move in a linear path unless there is an acceleration on the object. The acceleration for an object can be provided by a force acting towards the center of the path being followed by the object. This could be the tension of a string, electromagnetic force in the case of electrically charged objects, the force of gravity, etc.

If an object is rotating at 30 rpm and it is attached to a string that is R m long, the acceleration acting towards the center that maintains the path of the object is given by (2*pi/2)^2*R m/s^2

It can be seen that the acceleration acting on the object when it is 10 m away and rotating at 30 rpm is (2*pi/2)^2*10 m/s^2 and the acceleration when the object is 5 meters aways and rotating at 30 rpm is (2*pi/2)^2*5 m/s^2.

The work required to bring the object closer under the circumstances given is the definite integral of (2*pi/2)^2*R*M, where M is the mass of the object, from R = 10 to R = 5. This is equal to ((2*pi/2)^2*M/2)*(5^2 - 10^2) = (pi^2*M/2)*(25 - 100) = -37.5*M*pi^2 J

The work required in the last line is -37.5*M*pi^2 J. This is negative due to the fact that work is done by the object when the position changes instead of work having to be done on it for the same.

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