Question

A cylindrical rod of uniform density is located with its center at the origin, and its axis along the axis. Its radius is 0.03 m, its length is 0.7 m, and its mass is 4 kg. It makes one revolution every 0.05 s. If you stand on the axis and look toward the origin at the rod, the rod spins clockwise. What is the rotational angular momentum of the rod? What is the rotational kinetic energy of the rod? (Express your answer for rotational angular momentum in vector form.)

Answer 1

Answer:

angular momentum = 51.55 kg -m^2/s

rotational kinetic energy 3237.89 j

Explanation:

Given data;

radius = 0.03 m

length 0.7 m

mass is 4 kg

moment of inertia is given  $= \frac{1}{12} m (3r^2 + h^2)$

                       $I = \frac{1}{12} 10(3\times 0.03^2 + 0.7^2) = 0.4105 kg -m^2$

Angular speed$- \frac{2 \pi}{0.05} = 125.6 rad/s$

Angular momentum $= 125.6 \times 0.4105 = 51.55 kg- m^2/s$

Rotational kinetic energy $= 0.5 \times 0.4105 \times 125.6^2 = 3237.89 j$

Answer 2

The angular momentum of the rod is 51.5 kgm²/s and its rotational kinetic energy is 3234 J.

Given information:

Length of the cylindrical rod, l = 0.7 m

radius  of the cylindrical rod, r = 0.3 m

time period of revolution, T = 0.05s

mass of the rod, m = 4kg

Rotational motion:

From the time period we can calculate the angular speed (ω) of the rod:

$T =\frac{2\pi}{\omega}\\\\\omega=\frac{2\pi}{T}\\\\\omega= \frac{2\pi}{0.05} \\\\\omega=125.6\;rad/s$

The moment of inertia of the cylindrical rod is given by:

$I = \frac{1}{12}[3r^3+l^2]\\\\I = \frac{1}{12}[3(0.3)^3+(0.7)^2]\\\\I=0.41\;kgm^2$

So the angular momentum :

$L =I\omega\\\\L=0.41\times125.6\\\\L=51.5\;kgm^2/s$

The rotational kinetic energy of the rod:

$KE=\frac{1}{2}I\omega^2\\\\KE= \frac{1}{2}0.41\times(125.6)^2\\\\KE=3234\;J$

learn more about rotational motion:

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