Block with point mass rolling down [ Correlating variables and symmetry ]

Question

A symmetric block of mass 300 gm with a notch of the hemispherical shape of radius .2m rests on the smooth horizontal surface near the frictionless wall as shown in the figure. A point mass of 100gm slides from rest without friction from initial position B along with the notch. The maximum speed of the block is :

For attempting this question, I realized that when the point mass falls and climbs up to the top end of the rim from the depression in middle, it pushes on the block by the normal force and accelerates in the positive $x$ direction.

I did my free body diagram the following way:

Now the normal is given by $$ N= mg \cos \theta - ma_x \sin \theta$$ and from projecting it into orthogonal x-y components, I get the net force acting in the horizontal direction on the block using some trignometry:

$$ N + ma_{wedge} \sin\theta = mg \cos \theta$$\

$$ N = m ( g \cos \theta - a \sin \theta)$$

Equating normal to horizontal force on-ramp:

$$ Ma_x = m(g \cos \theta - a_x \sin \theta) \sin \theta$$

$$ a_x = \frac{mg \cos \theta \sin \theta}{M+m \sin^2 \theta}$$

$$ v_x = 2 \int \frac{mg \sin 2 \theta} { 2(M+m \sin^2 \theta) } dt$$

I thought of associating time to the motion of point mass:

$$ ds = r d \theta$$

$$ v_{p mass} = \sqrt{ 2gR} \sqrt{1 - \cos \theta}$$

$$ \frac{ r d \theta} { \sqrt{2gR ( 1 - \cos \theta) }} = dt$$

Plugging into integral,

$$ v_x = 2 \int_0^{ \frac{\pi}{2} } \frac{mg \sin 2 \theta}{2 ( M+ m \sin^2 \theta) } \frac{ R d \theta}{ \sqrt{2gR (1- \cos \theta) }}$$

$$ v_x = m \sqrt{\frac{gR}{2}} \int_0^{\frac{\pi}{2} } \frac{ \sin 2 \theta d \theta} { (M+ m \sin^2 \theta) \sqrt{1- \cos \theta} } $$

This integral doesn't converge itself I think.

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Now my questions are:

1. Is it correct to correlate time with the velocity of the block?
2. Is the amount of velocity imparted onto the block the same in the upward and downward motion of the block?
3. Are there any glaring mistakes in my work?

Answer 1

Note that you have two different motions, before and after the mass $m$ reaches the bottom. Before that point, the force from the small object is passed to the vertical wall. Only after that is the mass $M$ free to move.

Solving the problem is much easier if you understand the physics, and use conservation of energy. The minimum of the potential energy is at the bottom of the hemisphere. Let's call this $0$. Then the initial energy is $mgr$. After the mass $m$ goes pass the hemisphere minimum, it will start giving energy to the mass $M$ object, which will start to move. That means that object $m$ can't reach the same initial height, and it will start moving down the hemisphere. It will continue giving energy to the bigger object until it reaches an equilibrium at the bottom of the hemisphere, when both objects move at the same velocity.$$mgr=\frac12(M+m)v^2\\v=\sqrt{2gr\frac m{M+m}} \\v=\sqrt{2\cdot 10\cdot0.2\frac{100}{300+100}}=1$$

Answer 2

Here is my rather poor attempt at a diagram: Basically what I've done here is split the gravity force into components to get the normal force, then split the normal force into components to get the horizontal part of it, which is the part that pushes the ramp horizontally. Let the block have mass $m$ and the ramp have mass $M$. Let the ramp have radius $r$ and center initially at $(0,r)$. Let $x_1$ be the $x$ coordinate of the block and $x_2$ be the $x$ coordinate of the center of the ramp. Our little situation here is then described by the system of differential equations $$m\ddot{x}_1=mg\sin(\theta)(-\cos(\pi/2-\theta))\implies \ddot{x}_1=-mg\sin^2\theta$$ $$M\ddot{x}_2=mg\cos(\theta)\cos(\pi/2-\theta)\implies \ddot{x}_2=\frac{mg}{2M}\sin(2\theta)$$ Since the and assumedly the block reaches its initial position by first rolling down the left side of the ramp, and we assume the ramp starts at rest, we have the initial conditions $$x_1(0)=0 ~;~ \dot{x}_1=\sqrt{2gr} ~;~ x_2(0)=0 ~;~ \dot{x}_2(0)=0$$ Some calculus shows us that $$\tan(\theta)=\frac{x_1-x_2}{\sqrt{r^2-(x_1-x_2)^2}}$$ Thus, $$\theta =\begin{cases} \arctan\left(\frac{x_{1} -x_{2}}{\sqrt{r^{2} -( x_{1} -x_{2})^{2}}}\right) & x_{1} >x_{2}\\ \arctan\left(\frac{x_{1} -x_{2}}{\sqrt{r^{2} -( x_{1} -x_{2})^{2}}}\right) +\pi & x_{1} < x_{2} \end{cases}$$ This is obviously a nonlinear system and will not admit to closed form solutions. There are easier ways to reach the above using Lagrangian mechanics for example. If anyone finds any mistakes in the above please let me know - free body diagrams are notoriously confusing.