A solid disk, spinning counter-clockwise, has a mass of #16 kg# and a radius of #3/7 m#. If a point on the edge of the disk is moving at #8/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

The velocity at an angle is

where,

So,

The momentum of angular motion is

The angular momentum of the solid disk is $\frac{576}{35} \, \text{kg} \cdot \text{m}^2/\text{s}$.
The angular velocity of the disk is $\frac{40}{21} \, \text{rad/s}$.

The angular momentum ($L$) of the disk can be calculated using the formula:

$L = I \cdot \omega$

Where:

• $I$ is the moment of inertia of the disk.
• $\omega$ is the angular velocity of the disk.

The moment of inertia $I$ of a solid disk rotating about its center is given by:

$I = \frac{1}{2} m r^2$

Where:

• $m$ is the mass of the disk.
• $r$ is the radius of the disk.

Given:

• Mass ($m$) = 16 kg
• Radius ($r$) = $\frac{3}{7}$ m

Calculate $I$:

$I = \frac{1}{2} \times 16 \times \left(\frac{3}{7}\right)^2 = \frac{72}{49} \, \text{kg} \cdot \text{m}^2$

Given:

• Linear velocity ($v$) = $\frac{8}{5}$ m/s

The angular velocity $\omega$ can be calculated using the relationship between linear and angular velocity:

$v = r \cdot \omega$

Solve for $\omega$:

$\omega = \frac{v}{r} = \frac{\frac{8}{5}}{\frac{3}{7}} = \frac{56}{15} \, \text{rad/s}$

Now, calculate the angular momentum $L$:

$L = \frac{72}{49} \times \frac{56}{15} = \frac{3584}{735} \, \text{kg} \cdot \text{m}^2/\text{s}$

So, the angular momentum of the disk is $\frac{3584}{735} \, \text{kg} \cdot \text{m}^2/\text{s}$.

The angular velocity ($\omega$) of the disk is $\frac{56}{15} \, \text{rad/s}$.