Question

A sphere of uniform density with mass 22 kg and radius 0.7 m is spinning, making one complete revolution every 0.5 s. The center of mass of the sphere has a speed of4 m?s(a) What is the rotational kinetic energy Of the sphere? (b) What is the total kinetic energy of the sphere?

Short answer

(a) The rotational kinetic energy is, 340.46 J .

(b) The total kinetic energy is, 516.46 J.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of sphere is, V=4 m/s .
• The mass of sphere is, m=22kg
• The time of every revolution complete is, T=0.5 s
• The radius of the sphere is, r=0.7 m

Significance of total kinetic energy

The total kinetic energy of an object or system is equal to the sum of the kinetic energy from each type of motion.

(a): Determination of the rotational kinetic energy

The relation of rotational kinetic energy is expressed as,

$K_{rot}=\frac{1}{2}I{\omega }^2$ ...(i)

Here $K_{rot}$is the rotational kinetic energy, $\omega$is the angular speed and Iis the moment of inertia.

The moment of inertia is expressed as,

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

And the relation of angular speed is expressed as,

$\omega =\frac{2\pi }{T}$

Here $T$is the period of revolution.

Substitute the value of $I$and $T$in the equation (i).

$$\begin{gathered} K_{rot}=\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{2\pi }{T}\right)}^2 \\ =\frac{1}{5}m{r}^2{\left(\frac{2\pi }{T}\right)}^2 \end{gathered}$$

Substitute 0.5 s for T, 22 kg for m and 0.7 m for r in the above equation

$$\begin{gathered} K_{rot}=\frac{1}{5}\times 22\mathrm{kg}\times \left(0.7{\mathrm{m}}^2\right){\left(\frac{2\pi }{0.5\mathrm{s}}\right)}^2 \\ =340.46\mathrm{J} \end{gathered}$$

Hence the rotational kinetic energy is, 340.46 J.

(e): Determination of the total kinetic energy

The relation of the total kinetic energy is expressed as,

$K_{tot}=K_{trans}+K_{rot}$ ...(ii)

Here $K_{trans}$is the translational kinetic energy. It is expressed as,

$K_{trans}=\frac{1}{2}m{v}^2$

Substitute 4 m/s for v , and 22 kg for min the above equation.

$$\begin{gathered} K_{trans}=\frac{1}{2}\times 22\mathrm{kg}\times {\left(4\mathrm{m}/\mathrm{s}\right)}^2 \\ =176\mathrm{J} \end{gathered}$$

Substitute the value of $K_{trans}$and $K_{rot}$in the equation (ii).

$$\begin{gathered} K_{rot}=176\mathrm{J}+340.46\mathrm{J} \\ =516.46\mathrm{J} \end{gathered}$$

Hence the total kinetic energy is, $516.46\mathrm{J}$.