Question

A uniform sphere with a mass of 28.0 kg and a radius of 0.380 m is rotating at a constant angular velocity about a stationary axis that lies along the diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Short answer

Thus,The tangential velocity of a point on the rim of the sphere is$6.49\frac{m}{s}$.

Step-by-step solution

Step:-1 explanation

Given in the question, uniform sphere rotation about a fixed diameter.

$$\begin{gathered} M=28.0kg \\ R=0.380m \end{gathered}$$

Constant angular velocity$\omega$.

$K=236J$

Step:-2 calculation

We know that kinetic energy given by $K=\frac{1}{2}l{\omega }^2$

$\omega =\frac{v}{r}$, plug this equation into the expression for K.

$K=\frac{1}{2}l\frac{v^2}{R^2}$

The expression F and the moment of inertia I.

$$\begin{gathered} K=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{v^2}{R^2} \\ =\frac{1}{5}M{v}^2 \\ v=\sqrt{\frac{5K}{M}} \end{gathered}$$

Put the value $K=236$and $M=28.0$.

$$\begin{gathered} v=\sqrt{\frac{5\left(236\right)}{28.0}} \\ =\sqrt{\frac{1180}{28.0}} \\ =\sqrt{42.14} \\ =6.49\frac{m}{s} \end{gathered}$$

Hence , The tangential velocity of a point on the rim of the sphere is$6.49\frac{m}{s}$.