Question

Two disks are rotating about the same axis Disk A has a moment of inertia of 3.44 kg m2 and an angular velocity of 5.69 rad s Disk B is rotating with an angular velocity of 7.03 rad s The two disks are then linked together without the aid of any external torques so that they rotate as a single unit with an angular velocity of 4.23 rad s The axis of rotation for this unit is the same as that for the separate disks What is the moment of inertia of disk B

Answer 1

Answer:

The moment of inertia of disk B is 0.446 kilogram-square meters.

Explanation:

In this case, the moment of inertia of the disk B can be determined by means of the Principle of Conservation of Angular Momentum, whose model is:

$I_{A}\cdot \omega_{A} + I_{B}\cdot \omega_{B} = (I_{A} + I_{B})\cdot \omega$ (1)

Where:

$I_{A}$, $I_{B}$ - Moments of inertia of disks A and B, in kilogram-square meters.

$\omega_{A}$, $\omega_{B}$ - Initial angular velocities of disks A and B, in radians per second.

$\omega$ - Final angular velocity of the resulting system, in radians per second.

Let suppose that disk A rotates counterclockwise, whereas disk B rotates clockwise and that resulting system rotates counterclockwise. If we know that $I_{A} = 3.44\,kg\cdot m^{2}$, $\omega_{A} = 5.69\,\frac{rad}{s}$, $\omega_{B} = -7.03\,\frac{rad}{s}$ and $\omega = 4.23\,\frac{rad}{s}$, then the moment of inertia of the disk B is:

$I_{A} \cdot (\omega_{A} - \omega) = I_{B}\cdot (\omega - \omega_{B})$

$I_{B} = I_{A}\cdot \left(\frac{\omega_{A}-\omega}{\omega - \omega_{B}} \right)$

$I_{B} = (3.44\,kg\cdot m^{2})\cdot \left(\frac{5.69\,\frac{rad}{s} - 4.23\,\frac{rad}{s} }{4.23\,\frac{rad}{s} + 7.03\,\frac{rad}{s} } \right)$

$I_{B} = 0.446\,kg\cdot m^{2}$

The moment of inertia of disk B is 0.446 kilogram-square meters.