Question

1) What is the direction of the angular momentum for the rotation of the Earth?

2) If you give a hard boiled egg a spin on the table, it will keep spinning for a long time. If you do the same with an uncooked egg, it stops relatively quickly. Explain this.

3) A dumbbell consists of two uniform spheres of mass M and radius R, joined by a thin rod of radius r and length l and mass m.

a) What is the moment of inertia about an axis perpendicular to the rod, through the center of mass?

b) What is the moment of inertia about an axis parallel to the rod, through the center of mass?

4) A meterstick is initially standing vertically on a frictionless floor. A small gust of wind makes it begin to fall. At what angular velocity will it hit the floor?

Answer 1

1)

The angular momentum of the earth points along the axis of rotation of earth.

2)

for a boiled , the whole egg spins as one unit in the same direction and only things resisting the spin of the egg is the friction from the surface on which the egg is spinning.

In case of raw egg , there is liquids inside and solid cover outside. As the raw egg starts to sping sudenly from state of rest. due to inertia , the liquid inside spins in opposite direction and hence reduces the torque acting on the egg. since the torque is reduced , the raw egg spins for smaller period of time as compared to the boiled egg.

3)

a)

d = distance between center of sphere and axis = R + l/2

moment of inertia of each sphere about the axis using parallel axis theorem is given as

I_s = (0.4) MR² + M d² = (0.4) MR² + M (R + l/2)²

moment of inertia of rod is given as

I_r = ml²/12

net moment of inertia of dumbell will be

I_d = I_r + 2 I_s = ml²/12 + 2 ((0.4) MR² + M (R + l/2)²)

b)

I_rod = (0.5) m r²

I_s = moment of inertia of each sphere = (0.4) MR²

I_dumbell = I_rod + I_s = (0.5) m r² + (0.4) MR²

4.

L = length of stick = 1m

m = mass of stick

h = height of the center of mass = L/2 = 1/2 = 0.5 m

I = moment of inertia of meter stick = m L²/3

w = angular velocity

using conservation of energy

Rotational Kinetic energy = initial Potential energy

(0.5) I w² = mgh

(0.5) (m L²/3) w² = mgh

(0.5) (1²/3) w² = (9.8) (0.5)

w = 5.42 rad/s