A centrifuge has a rotational inertia of 5.50 $\times$ 10^-3 kg m².How much energy must be supplied to bring it from rest to 500 rad/s? A) 627 J B) 570 J C) 688 J D) 743 J E) 583 J

Question

Quizplus · Accepted Answer

The formula for rotational kinetic energy is:

$K = \frac{1}{2} I \omega^2$

where $I$ is the rotational inertia and $\omega$ is the angular velocity.

Plugging in the given values, we get:

$K = \frac{1}{2} (5.50 	imes 10^{10}	ext{ kg m}^2) (500	ext{ rad/s})^2 = 6.88 	imes 10^8	ext{ J}$.

However, the question asks how much energy must be supplied to bring the centrifuge from rest to 500 rad/s. This means that we need to find the difference between the initial and final kinetic energies, so we need to subtract the initial kinetic energy (which is zero) from the final kinetic energy:

$\Delta K = K_{final} - K_{initial} = K$

So the answer is:

$\Delta K = 6.88 	imes 10^8	ext{ J} - 0	ext{ J} = 6.88 	imes 10^8	ext{ J} \approx \boxed{	extbf{(C) } 688	ext{ J}}$.

A Centrifuge Has a Rotational Inertia of 5 ×\times× 10-3 Kg M2

A Centrifuge Has a Rotational Inertia of 5 $\times$ 10^-3 Kg M²