Question

Its mass of 1.5 m long, one end of which is tied to the ceiling of the room, is negligible, and a small object with a mass of 0.75 kg is hung on the free end of an inelastic rope. This object makes a circular motion at a speed of 25 rad / s, making an angle of 30 ° with the vertical.

a1) What is the rotational kinetic energy of this object?

a2) What is the rotational kinetic energy of the object if the angle is 60 °?

b) A thick stick of length L = 2.0m is denser on one end than the other, and the density is given by λ = (0.42 kg / m) - (0.15 kg / m2) x depending on the distance x to the dense end. The stick rotates around the axis that passes through the heavy end and is perpendicular to it in a period of T = 0.30s. Find the rotational kinetic energy of the stick.

Answer 1

Now since the object is rotating about the vertical axis tied with the rope and that the rope is making an angle of 30 with vertical thus, the radius around which the object will be rotating will be r = Lsin30 = 1.5sin30 = 1.5/2 = 0.75 m

Now the moment of inertia of the object around the axis will be

I = mr²

a1) Now the rotational kinetic energy is given as K_rot = 1/2 Iw²

=> K_rot = 1/2 x mr² x w² = 0.5 x 0.75 kg x (0.75m)² x 25² = 131.836 J

a2) r = Lsin60 = L/2 = 0.75 m

K_rot = 0.5 x 0.75kg x (0.75 m)² x 25² = 395.5 J

b) Given time period of the rotation thus angular speed w = 2/T = 20.944 rad/s

Now assume a point mass dm on the rod at a distance x from the dense end the moment of inertia of this mass can be given as dmx².

Now dm = λdx

=> dm = ((0.42 kg / m) - (0.15 kg / m2)x)dx

Thus moment of inertia of this point mass is ((0.42 kg / m) - (0.15 kg / m2)x)dx.x²

Now, integrating this for x = 0 to x = 2m we will get the total inertia of this rod about the axis;

I = ((0.42 kg / m)x² - (0.15 kg / m2)x³)dx

=> I = [0.42x³/3 - 0.15x⁴/4]₀²

=> I = 0.52 kgm²

Now the rotational kinetic energy of the rod about the axis can be given as

K_rot = 1/2 Iw² = 0.5 x 0.52kgm² x (0.0477s^-1)² = 114.05 J