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