You drill a hole at the end of a solid thin rod of length 2.8 m, and then attach

Question

You drill a hole at the end of a solid thin rod of length 2.8 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 10 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

A) for smallest angles

period is T = 2*pi*Sqrt(l/g) = 2*3.142*Sqrt(2.8/9.81) = 3.35 S

B) h = l*(1-cos(10)) = 2.8*(1-cos(10)) = 0.0425 m = 42.5 mm

C) m*g*h = 0.5*I*wmax^2

I is the moment of inertia = (1/3)*m*l^2

m*g*h = 0.5*(1/3)*m*L^2*wmax^2

m cancels on both sides

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

6*9.81*0.0425/(2.8*2.8) = wmax^2

wmax = 0.564 rad/s

at the T/4 = 3.35/4 =0.8375 s

linear speed is v = r*max = 2.8*0.564 = 1.58 m/s

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