# A sphere rolling down a rough wedge which lying on a smooth surface

A sphere of mass $m$ and radius $r$ rolls down from rest on an inclined (making an angle $\phi$ with the horizontal ) and rough surface of a wedge of mass $M$ which stays on a smooth horizontal floor. Find the acceleration of the wedge and the acceleration of the sphere relative to the wedge. I have written out the equations of motion for the sphere and the wedge and also the equation of motion for the rotational motion of the sphere. But then I get stuck. I don't know how I should relate the angular accleration $\alpha$ of the sphere with the acclerations $a_{\perp},a_{\parallel}$ appearing in its equations of motion. Can anyone please help me out? Thanks!

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Show us what have you done. – ja72 Mar 23 '14 at 15:39

Use Lagrangian mechanics method to answer this problem because it is easier than Newtonian mechanics (IMHO). Let $T$ be the kinetic energy, $V$ be the potential energy then the Lagrangian $L$ is given by $$L=T-V$$ and the Lagrangian equation is $$\frac{d}{dt}\left(\frac{dL}{d\dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0,$$ where it is assumed that $V$ is not a function of the velocities, i.e. $\dfrac{\partial v}{\partial\dot{q}_k}=0$. The symbol $q$ is a generalized coordinate used to represent an arbitrary coordinate $x, y,\theta$, etc and the 'dot' sign above $q$ means derivative with respect to time, $\;\dot{q}=\dfrac{dq}{dt}$.

Assuming the ball is uniform solid then its moment of inertia is $I=\dfrac{2}{5}mr^2$. Components of the velocity of the ball are: \begin{align} v_x&=\dot{x}+\dot{y}\cos\theta,\\ v_y&=\dot{y}\sin\theta,\\ v^2&=v_x^2+v_y^2=\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta. \end{align} The kinetic energies: \begin{align} T_{ball}&=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2\\ &=\frac{1}{2}mv^2+\frac{1}{2}\cdot\frac{2}{5}mr^2\omega^2\\ &=\frac{1}{2}mv^2+\frac{1}{5}mv^2\\ &=\frac{7}{10}mv^2.\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)\\ T_{wedge}&=\frac{1}{2}(M+m)\dot{x}^2.\\ T_{system}&=T_{ball}+T_{wedge}\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)+\frac{1}{2}(M+m)\dot{x}^2.\\ \end{align} The potential energy: $$V_{system}=V_{ball}=-mgy\sin\theta.$$ The Lagrangian is \begin{align} L&=T_{system}-V_{system}\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)+\frac{1}{2}(M+m)\dot{x}^2+mgy\sin\theta. \end{align} The Lagrange’s equations are $$\frac{d}{dt}\left(\frac{dL}{d\dot{x}}\right)-\frac{\partial L}{\partial x}=0$$ and $$\frac{d}{dt}\left(\frac{dL}{d\dot{y}}\right)-\frac{\partial L}{\partial y}=0.$$ After substituting the Lagrangian $L$ to the Lagrange’s equations, yield these two equations: $$\frac{7m}{5}\ddot{x}+\frac{7m}{5}\ddot{y}\cos\theta+(M+m)\ddot{x}=0\tag1$$ and $$\frac{7m}{5}\ddot{y}+\frac{7m}{5}\ddot{x}\cos\theta-mg\sin\theta=0.\tag2$$ Solving and simplifying $(1)$ and $(2)$ yield $$\ddot{x}=-\frac{5mg\sin\theta\cos\theta}{\left(5M+\left(5+7\sin^2\theta\right)m\right)}$$ and $$\ddot{y}=\frac{5(5M+12m)g\sin\theta}{7\left(5M+\left(5+7\sin^2\theta\right)m\right)},$$ where $\ddot{x}$ is the acceleration of the wedge and $\ddot{y}$ is the acceleration of the ball relative to the wedge.

$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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you have taken the translational energy of sphere twice. – user56199 Sep 25 '14 at 14:03

The constraint of motion of the two bodies is that the acceleration of the point of contact along the normal must be same otherwise the'll loose contact.

Get the acceleration of the point of contact of sphere along the normal equal to the acceleration vector of wedge along the normal.

The acceleration of the point of contact of sphere will include angular as well as linear acceleration and take it's component along common normal.

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