Constant rotational kinetic energy

fib1123 · Nov 20, 2017

A mass m is attached at the end of the string. The mass moves on a frictionless table, and the string passes through a hole in the table, under which someone is pulling on the string to make it taut at all times. Initially , the mass moves in a circle, with kinetic energy [E][/0] . The string is slowly pulled, until the radius of the circle is halved. How much work is done?

Diagram of a mass m on a frictionless table moving in a circle of radius r with speed v, tied to a string that passes through a hole in the table, with a downward force F pulling the string to a dog-like figure on the ground below.

I calculated the work and the result is

Handwritten derivation of work W = -∫ from R to R/2 of F·dr = l²/(2mr²) evaluated from R to R/2, giving l²/(2m)(4/R² - 1/R²) = 3l²/(2mR²).

but my question is: Why is the rotational kinetic energy constant during the motion?

A.T. · Nov 20, 2017

fib1123 said:

A mass m is attached at the end of the spring.

String?

fib1123 said:

Why is the rotational kinetic energy constant during the motion?

Work is done, so kinetic energy must change. The angular momentum is constant, because there is not torque.

Constant rotational kinetic energy

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