Question

Energy of 1000 J is spent to increase the angular speed of a wheel from 20 rad/s to 30 rad/s . Calculate it's moment of inertia.

Answer 1

Answer:0.8kg my

Explanation:

Answer 2

Moment of inertia of the wheel, $I=4\ kg-m^2$

Explanation:

Given that,

Energy spent, E = 900 J

Initial angular speed, $\omega_i=20\ rad/s$

Final angular speed, $\omega_f=30\ rad/s$

It is a case of conservation of energy in rotational kinematics. The change in kinetic energy is equal to the work done or the energy spent such that :

$\dfrac{1}{2}I\omega_f^2-\dfrac{1}{2}I\omega_i^2=W$

$\dfrac{1}{2}I(\omega_f^2-\omega_i^2)=W$

$I=\dfrac{2W}{(\omega_f^2-\omega_i^2)}\\\\I=\dfrac{2\times 1000}{((30)^2-(20)^2)}\\\\I=4\ kg-m^2$

So, the moment of inertia of the wheel is $4\ kg-m^2$. Hence, this is the required solution.

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