Question

A uniform solid sphere of radius r=0.500 m and mass m=15.0 kg turns counter-clockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is 3.00 rad/s.

Short answer

The angular momentum: $4.50kg\cdot m^2/s$

Step-by-step solution

Concept

The angular momentum is the amount of rotation of the body, which is the product of its inertia time and angular velocity.

(a) Vector angular momentum.

Angular momentum L is defined as the product of momentum of inertia I and angular speed $\text{ω}$.

$L=I\omega$

The momentum of inertia I of a solid sphere about its axis of rotation is given as follows:

$I=\frac{2}{5}m{r}^2$

Here, m is mass of sphere, and r is radius of sphere.

Substitute $\frac{2}{5}m{r}^2$for I in the equation for angular momentum and solve for L.

$\begin{array}{l}L=\left(\frac{2}{5}m{r}^2\right)\omega \\ L=\frac{2}{5}m{r}^2\omega \end{array}$

Substitute 15.0 Kg for m , 0.500m for r, and 3.00 rad/s forin the above equation and solve for L.

$\begin{array}{c}L=\frac{2}{5}(15.0k\text{Invalid <msup> element}g)(3.00rad/s)\\ =4.50kg\cdot m^2/s\end{array}$

Since the sphere rotates counterclockwise about the vertical axis y, the direction of angular momentum must be directed along positive z-axis.

Thus, the vector form of angular momentum is $4.50kg\cdot m^2/s$.

Result

The angular momentum: $4.50kg\cdot m^2/s$