Rolling Ball

$$ I use Lagrangian mechanics method to answer this problem because I think it is easier than Newtonian mechanics. Let $T$ be the kinetic energy, $V$ be the potential energy then the Lagrangian $L$ is given by $$ $L=T-V$ $$ and the Lagrangian equation is $$ $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0,$ $$ 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, $\;\dot{q}=\dfrac{dq}{dt}$ .

Let $M$ denote the wedge mass and $r$ denote the radius of the ball. Since 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 $$ $\begin{aligned} v_x&=\dot{x}+\dot{y}\cos\theta,\\ v_y&=\dot{y}\sin\theta,\\ v^2&=v_x^2+v_y^2=\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta. \end{aligned}$ $$ The kinetic energies: $$ $\begin{aligned} T_{ball}&=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2\\ &=\frac{1}{2}mv^2+\frac{1}{2}\cdot\frac{2}{5}mr^2\omega^2\\ &=\frac{1}{2}mv^2+\frac{1}{5}mv^2\\ &=\frac{7}{10}mv^2.\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)\\ T_{wedge}&=\frac{1}{2}(M+m)\dot{x}^2.\\ T_{system}&=T_{ball}+T_{wedge}\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)+\frac{1}{2}(M+m)\dot{x}^2.\\ \end{aligned}$ $$ The potential energy: $$ $V_{system}=V_{ball}=-mgy\sin\theta.$ $$ The Lagrangian is $$ $\begin{aligned} L&=T_{system}-V_{system}\\ &=\frac{7}{10}m\left(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos\theta\right)+\frac{1}{2}(M+m)\dot{x}^2+mgy\sin\theta. \end{aligned}$ $$ 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: $$ $\frac{7m}{5}\ddot{x}+\frac{7m}{5}\ddot{y}\cos\theta+(M+m)\ddot{x}=0$ $$ and $$ $\frac{7m}{5}\ddot{y}+\frac{7m}{5}\ddot{x}\cos\theta-mg\sin\theta=0.$ $$ Solving and simplifying the last two equations 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. Therefore, for $M=2m$ and $\theta=45^\circ$ , the acceleration of the ball is $$ $\begin{aligned} \ddot{y}&=\frac{5(5(2m)+12m)g\sin 45^\circ}{7\left(5(2m)+\left(5+7\sin^2 45^\circ\right)m\right)}\\ &=\frac{5(22)\cdot\frac{1}{2}\sqrt{2}g}{7\left(10+5+\frac{7}{2}\right)}\\ &=\frac{110}{259}\sqrt{2}g. \end{aligned}$ $$ Thus, $q-p=259-110=\boxed{149}$ . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

Because $v$ is the magnitude of the velocity of the ball with respect to the ground, and it rolls without slipping with respect to the wedge , the correct relation is $\dot{y}=\omega r$ .

There are several ways of doing the problem correctly, e.g., Newtonian mechanics with non-inertial frames, conservation of energy, or Lagrange equations, all yielding (obviously) the same correct answer $22$ . I will show the correct use of Lagrange's equation with his generalized coordinates $x$ and $y$ .

$T_{ball}=\frac{1}{2}m[(\dot{x}+\dot{y}\cos \theta)^2+(\dot{y}\sin \theta)]^2+\frac{1}{2}I\omega^2=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+2\dot{x}\dot{y}\cos \theta)+\frac{1}{5}m\dot{y}^2\\ T_{wedge}=\frac{1}{2}M\dot{x}^2\\ T=T_{ball}+T_{wedge}\\ V=-mgy \sin \theta$ Lagrange equation for $x$ and $y$ are $M\ddot{x}+m\ddot{x}+m\ddot{y}\cos\theta=0\\ \ddot{y}+\ddot{x}\cos\theta+\frac{2}{5}\ddot{y}-g\sin\theta=0$ Subsituting $\ddot{x}$ from the first equation and simplifying yields $\ddot{y}=\frac{5(m+M) \sin \theta}{7(m+M)-5 \cos^2 \theta}g$ Using $M=2m$ and $\theta =45^{\circ}$ , we find $\ddot{y}=\frac{15}{37}\sqrt{2}g$ So, $p=15,q=37\implies q-p = \boxed{22}$