Question

A cylindrical rod of uniform density is located with its center at the origin, and its axis along the x axis. It rotates about its center in the xy plane, making one revolution every 0.03 s. rod has a radius of 0.08 m, length of 0.7 m, and mass of 5 kg. It makes one revolution every 0.03 s. What is the rotational kinetic energy of the rod?

Short answer

The rotational kinetic energy is, 4653.35 J.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of cylindrical rod is, m = 5 kg .
• The radius of cylindrical rod is, r =0.08 m.
• The rotation period is, T =0.03 s.
• The length of cylindrical rod is, I =0.7 m .

Significance of rotational kinetic energy

The rotational kinetic energy is the form of energy that a moving object possesses through motion.

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 $I$is the moment of inertia.

The value of the moment of inertia and angular velocity for the disk is expressed as,

$I=\frac{1}{12}m{I}^2+\frac{1}{4}m{r}^2$

And

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

Here $m$is the mass of cylindrical rod, r is the radius of cylindrical rod and T is the rotation period.

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

$K_{rot}=\frac{1}{2}\left(\frac{1}{12}m{I}^2+\frac{1}{4}m{r}^2\right){\left(\frac{2\pi }{T}\right)}^2$

Substitute 5 kg for m, 0.08 m for r, 0.7 m for I , and 0.03 s for Tin the above equation.

$$\begin{gathered} K_{rot}=\frac{1}{2}\left(\frac{1}{12}\times 5\mathrm{kg}\times {\left(0.7\mathrm{m}\right)}^2+\frac{1}{4}\times 5\mathrm{kg}\times {\left(0.08\mathrm{m}\right)}^2\right){\left(\frac{2\pi }{0.03s}\right)}^2 \\ =4653.35\mathrm{J} \end{gathered}$$

Hence the rotational kinetic energy is,$4653.35\mathrm{J}$.