Oscillating Physical Pendulum The pendulum in Fig. 16-39 consists of a uniform d

Question

Oscillating Physical Pendulum The pendulum in Fig. 16-39 consists of a uniform disk with radius 10 cm and mass 550 g attached to a uniform rod with length 500 mm and mass 100 g

Oscillating Physical Pendulum The pendulum in Fig. 16-39 consists of a uniform disk with radius 10 cm and mass 550 g attached to a uniform rod with length 500 mm and mass 100 g Figure 16-39 (a) Calculate the rotational inertia of the pendulum about the pivot point. kg?m2 (b) What is the distance between the pivot point and the center of mass of the pendulum? m (c) Calculate the period of oscillation. s

Explanation / Answer

PART (A)
With the pivot point O about the upper end of the uniform rod and the uniform
disk connected at the other end of the rod, the total moment of inertia of the
system is
.......... Itot = Iro + Ido = (1/3)*Mr*Lr^2 + [ Idcm + Md*(Rd + Lr)^2 ]
(1) ..... Itot = (1/3)*Mr*Lr^2 + [ (1/2)*Md*Rd^2 + Md*(Rd + Lr)^2 ]
where the parallel axis theorem has been used to get the moment of inertia
Ido of the disk about the pivot point O. Substituting the values
Mr = 100 g = 0.100 kg , Lr = 500 mm = 0.5 m , Md =550 g = 0.550 kg ,
Rd = 10 cm = 0.10 m , we get Itot = 0.29 kg*m^2
ANSWER: 0.209 kg*m^2

PART (B)
The position of the center of mass relative to the pivot point O is given by
(2) ..... Ycm = [ Mr*(Lr/2) + Md*(Lr +Rd) ] / [ Mr + Md ]
Substituting the values, we get Ycm


PART (C)
The period of oscillation of a physical pendulum is given by
(2) ..... T = 2**sqrt [ I /(M*g* D ) ]
where D is the distance of the mass M from the axis of rotation and I is
the moment of inertia of the disk about the axis of rotation. Substituting
the values I = 0.209 kg*m^2 , M = 0.650 kg , and D = Ycm =

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