Finding the angular frequency of SHM of a rolling sphere

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SUMMARY

The discussion centers on calculating the angular frequency of a rolling sphere in simple harmonic motion (SHM) within a hemispherical bowl. The derived formula for angular frequency is sqrt(5g/(7(b-a))), where g is the acceleration due to gravity, b is the radius of the bowl, and a is the radius of the sphere. Key concepts include conservation of energy, kinetic energy (KE), potential energy (PE), and the moment of inertia of the sphere, which is I = 2/5 m (b-a)². The discussion emphasizes the importance of approximating small angles for accurate calculations.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with conservation of energy principles
  • Knowledge of kinetic and potential energy equations
  • Basic concepts of rotational dynamics, including moment of inertia
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  • Study the derivation of angular frequency in SHM scenarios
  • Explore the implications of small angle approximations in physics
  • Learn about the moment of inertia for various shapes and its applications
  • Investigate the relationship between potential energy and spring forces in SHM
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Students and educators in physics, particularly those focusing on mechanics and dynamics, as well as anyone interested in the principles of simple harmonic motion and rolling motion analysis.

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Homework Statement



A uniform sphere of radius a is placed at the lowest point of a fixed thin
hemispherical bowl of radius b > a. The sphere is the slightly displaced and
released with zero initial velocity such that it rolls without slipping on the inner
surface of the bowl. By conservation of energy, or otherwise, show that the
sphere executes simple harmonic motion about the lowest point with angular
frequency.

sqrt(5g/(7(b-a)))

Homework Equations



KE = 1/2 mv^2

Angular KE = 1/2 I [tex]\theta[/tex][tex]\dot{}[/tex][tex]^{2}[/tex]

dE/dt = 0

PE = mg ((b-a) - (b-a)cos([tex]\theta[/tex]))
= mg(b-a)(1 - cos([tex]\theta[/tex]))

I of sphere = 2/5 m (b-a)[tex]^{2}[/tex]

The Attempt at a Solution



KE = 1/2 m (b-a)[tex]^{2}[/tex] [tex]\theta[/tex][tex]\dot{}[/tex][tex]^{2}[/tex]

Total E = [tex]\theta[/tex][tex]\dot{}[/tex][tex]^{2}[/tex] (b-a)[tex]^{2}[/tex](1/2 + 1/5) + mg(b-a)(1 - cos([tex]\theta[/tex]))

dE/dt = 0 = [tex]\theta[/tex][tex]\ddot{}[/tex][tex]\theta[/tex][tex]\dot{}[/tex] (b-a)[tex]^{2}[/tex] * 7/5 + [tex]\theta[/tex] g(b-a)


I'm stuck with how to deal with having both theta dot and double dot in the first term, if I ignore the theta dot it works and I get the answer stated abov, but obviously something's gone wrong somewhere or else I'm not dealing with the thetas correctly!

Thanks
 
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The angle theta must be very small so you can replace cos theta with 1-theta^2/2.

You can consider theta as the coordinate and theta dot =omega as velocity. Look at the form of both the PE and KE. Like in case of SHM, PE is proportional to the square of coordinate, and KE is proportional to the square of velocity. They have the form PE=1/2*D (theta)^2, and KE=1/2 M (omega)^2. Find the equivalent mass M and the equivalent direction force (spring force) D.

ehild
 

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