- #1

#### Nitrus

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?

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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