- #1

vbrasic

- 73

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## Homework Statement

Find the acceleration of a uniform solid sphere (of mass ##m## and radius ##R##) rolling without slipping down an incline at angle ##\alpha## using the Lagrangian method.

## Homework Equations

Euler-Lagrange equation which says, $$\frac{\partial\mathcal{L}}{\partial q}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}},$$ where ##q## is some generalized coordinate.

## The Attempt at a Solution

A solid sphere rolling without slipping down an incline admits two types of kinetic energy. The first is associated with its rotation and is given by $$\frac{1}{2}I\dot{\theta}^2.$$ For a uniform solid sphere, $$I=\frac{2}{5}mR^2,$$ so the kinetic energy associated with rotation is $$\frac{1}{5}mR^2\dot{\theta}^2$$ The other type of kinetic energy it admits is the translational kinetic energy associated with the motion of the center of mass. Let the distance traveled by the com down the ramp be given by ##x##. We have that ##x=R\theta## (i.e. arc length). So, $$\dot{x}=R\dot{\theta}.$$ So, linear kinetic energy is given by $$\frac{1}{2}mR^2\dot{\theta}^2.$$ Overall then, $$T=\frac{3}{5}mR^2\dot{\theta}^2.$$ The potential energy is simply given by ##mgh##, where I define ##h## to be the height of the com. We have that the height is given by, $$h=-x\sin{\alpha}=-R\theta\sin{\alpha}.$$ So, the potential energy is $$-mgR\theta\sin{\alpha}.$$ So, the Lagrangian is, $$\mathcal{L}=\frac{3}{5}mR^2\dot{\theta}^2+mgR\theta\sin{\alpha}.$$ Then, using the Euler-Lagrange equation, we have that $$\frac{\partial\mathcal{L}}{\partial\theta}=mgR\sin{\alpha},$$ and, $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{6}{5}mR^2\ddot{\theta}.$$ We have, $$mgR\sin{\alpha}=\frac{6}{5}mR^2\ddot{\theta}.$$ Rearranging for ##\ddot{\theta}## gives $$\ddot{\theta}=\frac{5}{6}\frac{g}{R}\sin{\alpha}=\ddot{\theta}.$$ To find the linear acceleration, is all I need to do multiply this expression by ##R##?