Non-harmonic oscillatory motion

  • Thread starter Nitrus
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  • #1
I've got a test coming up with a problem similar to this one, I've figured out some of it but I am kinda lost on the rest, here it goes:
A solid sphere (radius = R) rolls without slipping in a cylindrical trough (radius = 5R). Show that, for small displacements from equilibrium perpendicular to the length of the trough, the sphere executes simple harmonic motion with a period T=2pi (28R/5g)^1/2.

Work:
I decided on taking an energy approach to this problem, and by doing so I must look at the KE of both the sphere and the effect the trough has on it.
[tex]
v= \frac {ds} {dt} = 4R \frac {d\theta} {dt}
[/tex]
[tex]
V=\frac {ds}{dt} = R\Omega
[/tex]
<p>
[tex]
\Omega =\frac {V} {R} = 4 \frac {d\theta} {dt}
[/tex]
with that we have the following (also including moment of intertia for the sphere)
[tex]
K = \frac {1} {2} 4R {\frac {d\theta}{dt}}^2 + \frac {1} {2}(\frac{2} {5} mR^2)(4{\frac {d\theta}{dt}}^2))
[/tex]
the trough is a half circle by the way...
that all simplifies to
[tex]
((\frac {d\theta}{dt}))^2 \frac {56mR^2}{5}
[/tex]
so now i have the energy of the system, what should i do next?
 
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Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
477
1
I don't know if you want your expression for kinetic energy checked (I didn't), but I can tell you that if you want to use an energy-based approach to this problem, you will also need an expression for the potential energy of the system. You can then write the Lagrangian and write out the equation of motion of the system. The period of oscillation will fall out of a small angle approximation of the equation of motion.
 
  • #3
krab
Science Advisor
893
3
Add the potential energy, which I believe is
[tex]4mgR(1-\cos\theta)\approx 2mgR\theta^2[/tex], note that energy is conserved, try harmonic variation for [tex]\theta[/tex], and you'll find your answer.
 

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