Question

its motion? 2) 3) An extremely light rod 1.00 m long has a 2.00-kg mass attached to one end and a 3.00-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 30.0 cm from the end with the 3.00-kg mass attached. The kinetic energy of the system is measured to be 100.0 J. (a) What is the moment of inertia of this system about the fixed axis? (b) What is the angular speed (in revolutions per second) of this system?

Answer 1

Answer:

a)   I = 1.25 Kg m² , b) w= 2.01 rev/s

Explanation:

a) in this exercise we are asked to find the moment of inertia of the system that is formed by the two masses, the bar has no mass because it is very light; let's use the definition of moment of inertia of a point mass

        I = m r₁² + M r₂²

                 

we use index 1 for the smallest mass m = 2kg and index 2 for the heaviest mass M = 3kg which indicates that the axis of rotation is at r₂ = 30 cm, so the other masses are at r₁ = 1- 0.3 = 0.7m from the axis of rotation

       I = 2 0.7² + 3 0.3²

       I = 0.98 +0.27

       I = 1.25 Kg m²

b) give us the kinetic energy of the system which is a scalar magnitude

      K = K₁ + K₂

       K = ½ I₁ w² + ½ I₂ w²

the angular velocity is the same for both because they are connected by the bar

        w² = 2K / (I₁ + I₂)

        W² = 2 100.0 / (0.98 + 0.27)

        w = √160

        w = 12.65 rad / s

reduce

         w = 12.65 rad/s (1rev/2pi rad)

          w= 2.01 rev/s