An aluminum ring rotates at 5 ms when an engine operates at

An aluminum ring rotates at 5 m/s when an engine operates at temperature of 150C, at which point the aluminum ring has a 5cm inner diameter and outer diameter of 5.5cm and 5mm in thickness. The engine is 25C when the engine is not running. When the engine starts, what is the rotational kinetic energy of the ring at the cold temperature?

Solution

Given:

v₁= 5 m/s--------------------T₁=150^oC

v₂=0 m/s--------------------T₂=25^oC

So ,from the above data we can say that velocity is directly proportional to Tempertaure

I _ring= m[r_out²- r _in²]

KE _rot= 1/2 *I _ring*w²=1/ 2*m [r_out²- r _in²]*w²=1/ 2*m [r_out²- r _in²]*[v²/r_out²]

KE _rot= 1/ 2*m [r_out²- r _in²]*[v²/r_out²] is the expression to fink rotational kinetic energy.