Question 625: Centre of Mass and Rotational Motion

Question

An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :

Accepted Answer

From work - energy theorem
$$W = \Delta k$$ (change in Kinetic Energy)
In rotation, $$KE = {1 \over 2}I{\omega ^2}$$
$$484 = {1 \over 2}I(\omega _f^2 - \omega _i^2)$$
$$ \Rightarrow 484 = {1 \over 2}I\left[ {{{\left( {2\pi {{360} \over {60}}} ight)}^2} - {{\left( {2\pi  	imes {{60} \over {60}}} ight)}^2}} ight]$$
$$ \Rightarrow 484 = {1 \over 2}I4{\pi ^2}(36 - 1)$$
$$ \Rightarrow I \simeq 0.7$$ kg-m
2

Answer

0.07 kg-m2

Answer

0.7 kg-m2

Answer

3.22 kg-m2

Answer

30.8 kg-m2

Centre of Mass and Rotational Motion

An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :

Choose the correct answer: