Question

A hollow hoop is hung on a pivot P, as shown in the figure, and oscillates back and forth. The radius of the hoop is R= 25.0 cm, and its mass is 3.0 kg.

A hollow hoop is hung on a pivot P, as shown in the figure, and oscillates back and forth. The radius of the hoop is R = 25.0 cm, and its mass is 3.0 kg. Using the equation derived in class for the period of a physical pendulum, calculate the hoop's period of oscillation in seconds. (Ic = MR2 for a hollow hoop.) Show all your work. A 2.0 kg point mass is attached to the hoop at the bottom point A in the diagram. Calculate the new period of oscillation. The answer may surprise you! (Hints: The total moment of inertia after the 2 kg is added will be the sum of the hoop's moment of inertia about P and the moment of inertia of the 2 kg point mass. The CM of the system moves to a distance d = (3.0 kg)R + (2.0 kg)(2R)/(3.0 + 2.0) kg = 7/5R below P.) Try a different value for the mass of the hoop and also for the mass added at A. (You can even try making the "added" mass negative, corresponding to a partial hoop with a piece cut out of it.) Calculate the new period of oscillation with and without the mass at A. What effect does changing the masses have on the period of oscillation?

Answer 1

Answer 2

Time period of a physical pendulum is given by T = 2*pi*sqrt(I/m*g*L)

where I is moment of inertia about hinge point and L is distance from hinge point to centre of mass

a)

Ic = mr^2

moment of inertia about hinge point is given by I = Ic+mr^2 = 2mr^2

=> T = 2*pi*sqrt(2mr^2/mgr) = 2pi*sqrt(2*r/g)

b)

when mass 'm1' is added at the bottom of hoop

I = 2mr^2 + m1*(2r)^2 = 2r^2(m+2m1)

total mass = m+m1 and L = 7/5r

=> T = 2*pi*sqrt(2r^2*(m+2m1)/((m+m1)*g*7/5r)) = 2*pi*((5r/7g)*(1+2m1/m)/(1+m1/m))