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 40 from vertical, so that it executes simple harmonic motion.
a.) What will be its oscillation period? 2.932 Correct: Your answer is correct. s
b.) When released, how high is the center of mass of the pendulum, above its lowest point during the oscillation? .243 Incorrect: Your answer is incorrect. mm
c.) What is the maximum angular speed of the pendulum during its oscillation?
max = 2.14 Incorrect: Your answer is incorrect. rad/s
and at what time does this first occur? .73 Correct: Your answer is correct. 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

given

L = 3.2 m

let m is the mass of the rod.

theta = 40 degrees

moment of inertia of rod about one end, I = m*L^2/3

a) Time period of physical pendulum,

T = 2*pi*sqrt(I/(m*g*Lcm)

= 2*pi*sqrt(m*L^2/3/(m*g*L/2))

= 2*pi*sqrt(2*L/(3*g))

= 2*pi*sqrt(2*3.2/(3*9.8))

= 2.932 s

b) h = (L/2)*(1 - cos(40))

= (3.2/2)*(1 - cos(40))

= 0.374 m

c) Apply conservation of energy

rotationalkinetic energy of rod when it is vertical = initial potentail energy of the rod

(1/2)*I*wmax^2 = m*g*h

(1/2)*(m*L^2/3)*wmax^2 = m*g*h

(L^2/6)*wmax^2 = g*h

wmax^2 = 6*g*h/L^2

wmax = sqr(6*g*h)/L

= sqrt(6*9.8*0.374)/3.2

= 1.465 rad/s

d) Vmax = (L/2)*wmax (using v = r*w)

= (3.2/2)*1.465

= 2.344 m/s

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