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

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;

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

is the length of the rod ( 2.6 m )

so we substitute

× 3 × (2.6)²

=   1.69 kg.m²

now, to calculate the rotational kinetic energy,

kinetic energy due to rotation is;

KE =

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

w is angular frequency ( 37 radians/s )

we substitute

KE = × 1.69 × ( 37 radians/s )²

KE = 1156.81 J

Therefore, rotational kinetic energy is 1156.81 J