# D'Alemberts Principle to solve a Classical Mechanics problem

I found this question on this very site and was curious on how to solve it using D'Alembert's Principle

.I already know how to do it by balancing the torque about toppling point so please don't post the latter as an answer .

My attempt was to displace the upper sphere by $$d\theta$$ which gives its virtual displacement as

$$rd{\theta}cos(\theta)$$j(notice we are only interested in the y component becuz the only component of the applied force which is gravity in this case is the y component in the dot product the other components vanish)

similarly the cylinder is also displaced $$r'd{\phi}cos(\phi)$$ where r' is the distance from the point of toppling of the center of mass.

$$r'cos(\phi)=H/2=r(1+sin(\theta))$$

,$$r'sin(\phi)=R=r(1+cos(\theta))$$ and

$$tan(\phi)=H/2R$$

the equation coming from D'Alemberts principle is $$mgrcos(\theta)d\theta=Mgr'cos(\phi)\phi$$

I tried solving this but couldn't arrive at the right answer.

Any help would be appreciated!

• – Urb Nov 10 '20 at 23:25
• Thats where I got the question from.I didn't understand how to solve using D'Alembert's principle and hence the post – padma bilgi Nov 11 '20 at 5:30

Suppose the line joining the centres of the 2 spheres makes an angle $$\theta$$ with the vertical. Suppose this line rotates through small angle $$\delta \theta$$ clockwise as the upper sphere rotates around the rim of the lower sphere. The CM of the upper sphere moves a distance $$2r\delta \theta$$ perpendicular to the line joining the centres, where $$r$$ is the radius of the spheres. So the CM moves horizontally to the right by $$\delta x=2r\cos\theta \delta \theta$$ and moves vertically down by $$\delta y=2r\sin\theta \delta \theta$$.
In order to allow the upper sphere to move horizontally, the cylinder must tilt through an angle $$\delta \phi$$ about its point of contact with the lower sphere. The difference in heights at which the upper and lower spheres touch the cylinder is $$b=2r\cos\theta$$. The horizontal displacements of the cylinder and upper sphere must be equal : $$\delta x=b\delta \phi$$.
The CM of the cylinder lies on its axis. As the cylinder tilts through $$\delta \phi$$, the CM rises by distance $$R\delta\phi$$ where $$R$$ is the radius of the cylinder.
Note that $$2r\sin\theta=2R-2r$$ so $$r\sin\theta=R-r$$.
Pulling everything together, and applying the Principle of Virtual Work : $$mg\delta y=MgR\delta \phi$$ $$2mr\sin\theta \delta \theta=MR\delta\phi=MR\frac{\delta x}{b}=MR\frac{2r\cos\theta\delta \theta}{2r\cos\theta}=MR\delta\theta$$ $$2m(R-r) =MR$$