You drill a hole at the end of a solid thin rod of length 3.2 m, and then attach

Question

You drill a hole at the end of a solid thin rod of length 3.2 m, and then attach the rod through this hole onto an approximately frictionless pivot. At t=0, you release the rod from rest at a small angle of 1 degree from vertical, so that it executes simple harmonic motion.

a.) What will be its oscillation period?
  s

b.) When released, how high is the center of mass of the pendulum, above its lowest point during the oscillation?
  mm

c.) What is the maximum angular speed of the pendulum during its oscillation?
max =   rad/s
and at what time does this first occur?   s

Also, find the maximum linear speeds of:
the pendulum's center of mass:   m/s
the lowest point of the pendulum:   m/s

Explanation / Answer

T = 2pi*sqrt(I/mgd)

here I = 1/3ml^2

d = l/2

T = 2*3.14*sqrt(1/3ml^2/mgl/2)

T = 6.28sqrt(2/3*3./9.8) = 2.93 sec

2) h = l*(1-costheta) = 3.2*(1-cos1) = 0.0032 m = 3.2 mm

3) mgh = 1/2Iw^2

i = 1/3ml^2

mgh = *1/2*1/3ml^2*w^2

w = sqrt(6gh/l^2) = 0.1355 rad/sec

w = 2pi/T

T = 2*3.14/w = 6.28/0.1355 = 46.35 sec

v = rw = l/2*w = 0.217 m/sec

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