Question

PROBLEM 2.In the sketch a three part physical pendulum is shown, consisting of two massless rods (L=1 m) which make a 90 angle with respect to each other and are constrained to pivot at about an axis that is perpendicular to the paper and at the corner of where they meet. Two unequal point masses aresolidly attached, one at each end. The rod oscillates (when disturbed from equilibrium) due to the downward force of gravity. (ignore friction and air resistance). m1 and m2 are point masses., m=1kg

a.What is the rotational inertia of this pendulum

b.Where is the system Center of Mass? (With respect to the pivot point)

c.What is the period of oscillation.!d.If the two point masses were spheres instead, but still centered at the ends of the rod, would the period of the pendulum change and if so would it increase or decrease? (Justify your answer conceptually)

Answer 1

a)I = (1/2) x (m1 + m2) x r^2

     =0.5( mr^2+mr^2)

   I = 0.5(1+1) = 1

b)sqrt of (square of the distances)

   = sq rt(L^2/4+L^2/4)

   = L/1.414

   = 1/1.414

   = 0.707

c)T = 2pi x sqrt (L/g)

T = 2* 3.14*sqrt(1/9.8)

     = 2.007

d) no change becuase there is no change in center of mass

Answer 2

the inertia is

I = (1/2) x (m1 + m2) x r^2

the period of oscillation is

T = 2pi x (L/g)^1/2

where L is length and g = 9.8 m/s^2