Complete Question:
A uniform rod of mass 1.90 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 = 5.40 kgis attached to one end and a second mass m2 = 2.50 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 2.70 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 2.70 rad/s?
Answer:
a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J
Explanation:
a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.
As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.
The total rotational inertia will be as follows:
I = M*L²/12 + m₁*r₁² + m₂*r₂²
⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²
⇒ I = 8.53 kg*m²
b) The rotational kinetic energy of the rigid body composed by the rod and the point masses m₁ and m₂, can be expressed as follows:
Krot = 1/2*I*ω²
if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:
Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²
⇒ Krot = 31.1 J
c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:
I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²
I = 7.90 kg*m²
d) The new rotational kinetic energy will be as follows:
Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²
Krot= 28.8 J
