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## Homework Statement

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A uniform solid ball of mass [itex]m[/itex] rolls without slipping down a right angled wedge of mass [itex]M[/itex] and angle [itex]θ[/itex] from the horizontal, which itself can slide without friction on a horizontal floor. Find the acceleration of the ball relative to the wedge.

**2. The attempt at a solution**

Attempt using generalized coordinates:

Let us denote the coordinate of the ball as [itex]ξ[/itex] (which is along the incline) and the coordinate of the wedge as [itex]η[/itex]. The kinetic energy of the system is:

[tex] K=\frac{1}{2}m\dotξ^2+\frac{1}{2}Iω^2+\frac{1}{2}M\dotη^2=\frac{1}{2}m\dotξ^2+\frac{2}{10}m\dotξ^2+\frac{1}{2}M\dotη^2=\frac{7}{10}m\dotξ^2+\frac{1}{2}M\dotη^2 [/tex]

And the change in potential energy is:

[tex] V=-mgξ\sin{θ} [/tex]

The sum of both energies is the total energy, which is constant in time. We take the derivative of it:

[tex] \dot{E}=\frac{d}{dt}E=\frac{d}{dt} K+V=\frac{d}{dt} (\frac{7}{10}m\dotξ^2+\frac{1}{2}M\dotη^2-mgξ\sin{θ})=\frac{7}{5}m\dotξ\ddotξ+M\dotη\ddotη-mg\dotξ\sin{θ}=0 [/tex]

Finally, we need to relate both coordinates. Since the center of mass of the system doesn't move we have:

[tex]mξ\cos(θ)=Mη[/tex] [tex]\frac{mξ\cos(θ)}{M}=η[/tex]

We substitute [itex]η[/itex] in our derivative and we get:

[tex]\dot{E}=\frac{7}{5}m\dotξ\ddotξ+\frac{m^2\dot{ξ}\ddot{ξ}\cos^2{θ}}{M}-mg\dot{ξ}\sin{θ}=0 [/tex]

We solve for [itex]\ddotξ[/itex]:

[tex]\ddotξ =\frac{5Mg\sin{θ}}{7M+5m\cos^2{θ}}[/tex]

Can someone please confirm if this solution is correct?

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