1 Answer | Add Yours
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.
We’ve answered 317,573 questions. We can answer yours, too.Ask a question