Question

A 1.50-kg rod 1.00m long has a 2.00-kg mass attached to one end and a 4.00-kg mass attached to the other. The system rotates at a constant angular speed about a fixed axis perpendicular to the rod that passes through the rod 40.0 cm from the end with the 4.00-kg mass attached. The angular speed of the system is 150 rad/s.
a) What is the total moment of inertia of this system (including the rod and two masses) about the fixed axis?

b) What is the kinetic energy of this system (including the rod and two masses)?

Answer 1

here,

mass of rod , m = 1.5 kg

m1 = 2 kg

m2 = 4 kg

length , l = 1 m

a)

the moment of inertia of the system about an axis passing through its center , I1 = (m1 * (l/2)^2 + m2 *(l/2)^2 + m * l^2 /12)

using parallel axis theorm

the moment of inertia about a point that passes through 40 cm mark , I = I1 + (m + m2 + m2) * (l/2 - 0.4)^2

I = (m1 * (l/2)^2 + m2 *(l/2)^2 + m * l^2 /12)+ (m + m2 + m2) * (l/2 - 0.4)^2

I = (2 * (1/2)^2 + 4 *(1/2)^2 + 1.5 * 1^2 /12)+ (1.5 + 2 + 4) * (1/2 - 0.4)^2

I = 1.7 kg.m^2

b)

angular speed ,w = 150 rad/s

the kinetic energy of the system , Ke = 0.5 * I * w^2

Ke = 0.5 * 1.7 * 150^2 = 1.91 * 10^4 J