Question

A thin uniform-density rod whose mass is 3.0 kg and whose length is 2.6 m rotates around an axis perpendicular to the rod, with angular speed 37 radians/s. Its center moves with a speed of 13 m/s. (a) What is its rotational kinetic energy

Answer 1

Answer:

rotational kinetic energy is 1156.81 J

Explanation:

Given the data in the question;

first we find the moment of inertia of the rod as axis passes through the middle or center;

$I = \frac{1}{12}ml^2$

where m is the mass of the road( 3.0 kg)

$l$ is the length of the rod ( 2.6 m )

so we substitute

$I = \frac{1}{12}$ × 3 × (2.6)²

$I$ =   1.69 kg.m²

now, to calculate the rotational kinetic energy,

kinetic energy due to rotation is;

KE$_R$ = $\frac{1}{2}Iw^2$

where $I$ is moment of inertia( 1.69 kg.m² ),

w is angular frequency ( 37 radians/s )

we substitute

KE$_R$ = $\frac{1}{2}$ × 1.69 × ( 37 radians/s )²

KE$_R$ = 1156.81 J

Therefore, rotational kinetic energy is 1156.81 J

Answer 2

The rotational kinetic energy will be 1156.81 J. It is half of the product of the moment of inertia and angular speed.

What is rotational kinetic energy?

The rotational kinetic energy is the kinetic energy generated by an object's rotation and It is a component of its overall kinetic energy.

The given data in the problem is;

m is the mass = 3.0 kg

l is the length = 2.6 m

$\rm \omega$ is the angular speed = 37 radians/s

v is the speed = 13 m/s

$\rm E_k$ is the  rotational kinetic energy=?

The momentum of inertia is found by;

$\rm I = \frac{1}{12} ml^2 \\\\ \rm I = \frac{1}{12} \times 3 \times (2.6)^2 \\\\\ \rm I = 1.69 \kg.m^2$

The rotational kinetic energy is found by;

$\rm E_k= \frac{1}{2}I (\omega)^2 \\\\ \rm E_k= \frac{1}{2}\times 1.69 (37)^2 \\\\ \rm E_k=1156.81 \ J$

Hence the rotational kinetic energy will be 1156.81 J.

To learn more about the rotational kinetic energy refer to the link;

https://brainly.com/question/19305456