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Wize AP Physics C: Mechanics Textbook > Unit 6: Oscillations (6-14%)

Physical Pendulum

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Physical Pendulum


If the mass is not concentrated at the lower part of the object (like a simple pendulum (mass on a weightless string)), still the object can oscillate around its pivot point.
  • This oscillation for small angles could be also described by simple harmonic motion.
  • For this oscillation:
θ(t)=θmaxcos⁡(ωt+ϕ0)\theta(t)=\theta_{max}\cos (\omega t+ \phi_0)
Where

ω=MglI\boxed{\omega=\sqrt{\frac{Mgl}{I}}}

Where
  • II is the moment of inertia about the pivot point. (Don’t worry about formulas of moment of inertia. If you need it in exam, it should be given to you in the formula sheet.)
  • ll is the distance between the pivot point and the center of mass of the object.

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A rod with length of 20 cm and the mass of 0.5kg is hanging from one end of it and it is set oscillating. The mass distribution along the rod is uniform and the moment of inertia for a uniform rod of length d respect to one end of it is 13Md2\frac{1}{3}Md^2. Find the time it takes for this physical pendulum to go from maximum displacement to minimum displacement.

Solution:
Time time to go from Max displacement (x=+A) to minimum displacement (x=-A) is half of one period. Hence, that is enough to find the period of this oscillation first.

ω=MglI\omega=\sqrt{\frac{Mgl}{I}} where llis the distance form the pivot point to the centre of mass. Since we have a uniform rod, its centre of mass would be at the middle of the rod and since it is hanging from one end, the distance between the pivot point and the centre of mass is: l=(d/2)=0.22=0.1ml=(d/2)=\frac{0.2}{2}=0.1m

Thus: ω=Mgl13M(2l)2=3g4l=3(9.8)4(0.1)=8.57rad/s\omega=\sqrt{}\frac{Mgl}{\frac{1}{3}M(2l)^2}= \sqrt{\frac{3g}{4l}}=\sqrt{\frac{3(9.8)}{4(0.1)}}=8.57 rad/swhere I used the fact that d=2ld=2l
From this: T=2πω=0.733sT=\frac{2\pi}{\omega}=0.733sand finally the time we are looking for is: t=T2=0.36644st=\frac{T}{2}=0.36644s
You have landed on an unknown planet. You have a simple pendulum with a length of 25.0 cm25.0\ cm. You measure the frequency of oscillation of the pendulum to be 0.5 Hz0.5\ Hz.

a) What is the gravitational acceleration on this planet? (in m/s^2)
b) What would be the period of the oscillation if you used this pendulum on Earth? (in seconds)

You pull a simple pendulum to the side and release the mass at an angle of 2.23°2.23\degree.

If the length of the pendulum is 25 cm, how much time does it take the mass to reach its highest speed? (in s)
The position of an oscillating mass in a simple pendulum is given as: x(t)=3.56cos⁡(12t+π)x(t)=3.56\cos(12t+\pi).
What is the maximum velocity of the mass?