Question

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.44 kg and rotate with the same angular speed of 430 rad/s, but they differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius 0.356 m, and (b) the larger cylinder, of radius 0.775 m?

Answer 1

Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

$I= \frac{1}{2}mr^{2}$

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

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

Here, K is the rotational kinetic energy and $\omega$ is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.356 m.

$I= \frac{1}{2}(3.44)(0.356)^{2}$

$I= 0.218 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

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

Put $I= 0.218 kg m^{2}$ and $\omega = 430 rads^{-1}$.

$K= \frac{1}{2}(0.218)(430) ^{2}$

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.775 m.

$I= \frac{1}{2}(3.44)(0.775)^{2}$

$I= 1.03 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

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

Put $I= 1.03 kg m^{2}$ and $\omega = 430 rads^{-1}$.

$K= \frac{1}{2}(1.03)(430) ^{2}$

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.