Question

A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.30 kg is attached to one end and a second mass m2 = 3.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
A) What is the moment of inertia of the system?B) If the rod rotates with an angular speed of 2.00 rad/s, how much kinetic energy does the system have?C) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?D) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.00 rad/s?

Answer 1

Answer:

Explanation:

Moment of inertia of the rod = 1/12 m L²

m is mass of the rod and L is its length

= 1/2 x 2.3 x 2 x 2

= 4.6 kg m²

Moment of inertia of masses attached with the rod

= m₁ d² + m₂ d²

m₁ and m₂ are masses attached , and d is their distance from the axis of rotation

= 5.3 x 1² + 3.5 x 1²

= 8.8 kg m²

Total moment of inertia = 13.4 kg m²

B )

Rotational kinetic energy = 1/2 I ω²

I is total moment of inertia and ω is angular velocity

= .5 x 13.4 x 2²

= 26.8 J .

C )

when mass of rod is negligible , moment of inertia will be due to masses only

Total moment of inertia of masses

= 8.8 kg m²

D )

kinetic energy of the system

= .5 x 8.8 x 2²

= 17.6 J .