What is the maximum displacement of the wagon during the motion of the bead?

[Karol]

Homework Statement

The bead of mass m starts to slide without friction from point A. the mass of the cart and the rail is M.
What is the max displacement of the wagon during the motion of the bead?
What is the force the rail exerts on the cart when the cart's velocity is maximum?
What is the impulse exerted on the bead from the moment it passes for the first time at point B until it passes the first time at point C?

Homework Equations

Impulse and momentum: Ft=mv

The Attempt at a Solution

Relative to the origin at the center of the arc, point O, the COM, at the beginning, is:
$$x_cm=\frac{-mR}{m+M}$$
The bead moves to the other side and the COM moves to a distance x from the origin:
$$\frac{-mR}{m+M}=\frac{mR-xM}{m+M}\Rightarrow x=\frac{2rm}{M}$$
The displacement of the wagon is:
$$\Delta x=\frac{2rm}{M}-\frac{mR}{m+M}$$
The force the rail exerts on the cart when the cart's velocity is at the max is zero since then the bead is at the lowest position (point C), it's velocity is horizontal and it doesn't exert any force on the rail.
The impulse from point B to C:
$$Ft=mv\rightarrow Ft=m(v_c-v_b)$$

 

[jbriggs444]

The force the rail exerts on the cart when the cart's velocity is at the max is zero since then the bead is at the lowest position (point C), it's velocity is horizontal and it doesn't exert any force on the rail.

Its velocity is horizontal. But Newton's second law does not say anything about velocity, does it?

 

[Orodruin]

Relative to the origin at the center of the arc, point O, the COM, at the beginning, is:
$$x_cm=\frac{-mR}{m+M}$$
The bead moves to the other side and the COM moves to a distance x from the origin:
$$\frac{-mR}{m+M}=\frac{mR-xM}{m+M}\Rightarrow x=\frac{2rm}{M}$$
The displacement of the wagon is:
$$\Delta x=\frac{2rm}{M}-\frac{mR}{m+M}$$

The bead does not move from $-mR$ to $mR$ since the rail moves with the cart and the cart is moving.

The force the rail exerts on the cart when the cart's velocity is at the max is zero since then the bead is at the lowest position (point C), it's velocity is horizontal and it doesn't exert any force on the rail.

Force is not proportional to velocity, it is proportional to acceleration and the acceleration at the lowest point is not zero.

The impulse from point B to C:
$$Ft=mv\rightarrow Ft=m(v_c-v_b)$$

This does not really answer the question. You need to find the difference in velocities. (Note that velocity is a vector quantity!)

 

[andrewkirk]

It might be easiest to look at the vertical and horizontal components of the impulse, then vector add them.

The vertical component is easy: it's pointing upwards and has magnitude equal to the downwards momentum of the bead at B.

To work out the horizontal component I think you'll need to use conservation of both energy and momentum. Let $v$ be the bead's velocity relative to the ground (not the cart) at C Conservation of momentum will then give you the speed of the cart in terms of $v, m, M$. You can then work out the total kinetic energy of bead plus cart when bead is at C, in terms of those variables. Equate that kinetic energy to $mh$ and solve for $v$. Then you can work out the bead's horizontal momentum at C, which will equal the horizontal impulse from B to C.

 

[jbriggs444]

It might be easiest to look at the vertical and horizontal components of the impulse, then vector add them.

The vertical component is easy: it's pointing upwards and has magnitude equal to the downwards momentum of the bead at B.

To work out the horizontal component I think you'll need to use conservation of both energy and momentum. Let $v$ be the bead's velocity relative to the ground (not the cart) at C Conservation of momentum will then give you the speed of the cart in terms of $v, m, M$. You can then work out the total kinetic energy of bead plus cart when bead is at C, in terms of those variables. Equate that kinetic energy to $mh$ and solve for $v$. Then you can work out the bead's horizontal momentum at C, which will equal the horizontal impulse from B to C.

I agree with the approach but would quibble about the total kinetic energy when the bead is at point C (bottom point on the rail). That should include a contribution from R.

 

[andrewkirk]

the total kinetic energy when the bead is at point C (bottom point on the rail). That should include a contribution from R.

What did you have in mind as a contribution from R? My interpretation of the diagram is that R is a radius, not an object.
Or did you interpret R as referring to the rail? By 'cart' I am referring to the cart+rail, which has mass M.

 

[Orodruin]

What did you have in mind as a contribution from R?

I believe he is referring to the fact that you quoted the lost potential energy (and thus the total kinetic energy) as mh and not m(h+R). When the bead is at the bottom, it is a vertical distance h+R from where it started.

 

[andrewkirk]

Ah yes. Quite right.

 

[Karol]

The bead does not move from $-mR$ to $mR$ since the rail moves with the cart and the cart is moving.

$$\Delta x=\frac{m(R-x)-Mx}{m+M}-\frac{-mR}{m+M}=\frac{2mR-(m+M)x}{m+M}$$

Force is not proportional to velocity, it is proportional to acceleration and the acceleration at the lowest point is not zero.

The rail exerts the centripetal force on the wagon. conservation of energy and momentum:
$$\left\{ \begin{array}{l} 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=MV_c \end{array}\right.\Rightarrow v_c^2=\frac{4MG(h+R)}{m(m+M)}$$
$$F_{centripetal}=m\frac{v^2}{R}=m\frac{4MG(h+R)}{Rm(m+M)}=\frac{4MG(h+R)}{R(m+M)}$$

This does not really answer the question. You need to find the difference in velocities. (Note that velocity is a vector quantity!)

$v_b^2=2gh$. v_b points downward, v_c is horizontal
$$\sqrt{v_c+v_b}=\sqrt{2gh+\frac{4MG(h+R)}{m(m+M)}}$$
$$\vec{F}t=m\vec{v}\rightarrow \vec{F}t=m(\vec{v_c}-\vec{v_b})=m(\sqrt{2gh+\frac{4MG(h+R)}{m(m+M)}})$$
The impact's angle is:
$$\tan\alpha=\frac{v_c}{v_b}$$
The impact acts upward and to the right

 

[Orodruin]

$$\Delta x=\frac{m(R-x)-Mx}{m+M}-\frac{-mR}{m+M}=\frac{2mR-(m+M)x}{m+M}$$

Now you have introduced two variables $x$ and $\Delta x$. Originally you specified that $x$ was the displacement of the cart. What is $\Delta x$?

The rail exerts the centripetal force on the wagon. conservation of energy and momentum:
$$\left\{ \begin{array}{l} 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=MV_c \end{array}\right.\Rightarrow v_c^2=\frac{4MG(h+R)}{m(m+M)}$$

You have a dimensional error here. Your velocity squared has units of length^2 / (time^2 mass). Think about where this may have entered.

$$F_{centripetal}=m\frac{v^2}{R}=m\frac{4MG(h+R)}{Rm(m+M)}=\frac{4MG(h+R)}{R(m+M)}$$

The last steps would be correct if the cart was stationary, but the cart is moving. Think about which velocity you need to put into $mv^2/R$.

$v_b^2=2gh$. v_b points downward, v_c is horizontal
$$\sqrt{v_c+v_b}=\sqrt{2gh+\frac{4MG(h+R)}{m(m+M)}}$$

You hare missing squares here, which had me confused for some time, but you have used it as if you had squared the velocities. It would be correct if the expression for the velocity at C was correct.

 

[Karol]

Now you have introduced two variables $x$ and $\Delta x$. Originally you specified that $x$ was the displacement of the cart. What is $\Delta x$?

$$\frac{m(R-x)-Mx}{m+M}=\frac{-mR}{m+M}\Rightarrow x=\frac{2mR}{m+M}$$

You have a dimensional error here. Your velocity squared has units of length^2 / (time^2 mass). Think about where this may have entered.

I don't find the error:
$$\left\{ \begin{array}{l} 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=MV_c\rightarrow V_c=\frac{m}{M}V_c \end{array}\right.\Rightarrow 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}M\frac{m^2}{M^2}v_c^2$$
$$\rightarrow 2g(h+g)=\frac{1}{2}mv_c^2+\frac{1}{2}m\left( 1+\frac{m^2}{M} \right)v_c^2$$
It isn't correct.

The last steps would be correct if the cart was stationary, but the cart is moving. Think about which velocity you need to put into $mv^2/R$.

$$F_{centripetal}=m\frac{(v_c+V_c)^2}{R}$$

 

[Orodruin]

$$\frac{m(R-x)-Mx}{m+M}=\frac{-mR}{m+M}\Rightarrow x=\frac{2mR}{m+M}$$

Correct.

I don't find the error:
$$\left\{ \begin{array}{l} 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=MV_c\rightarrow V_c=\frac{m}{M}V_c \end{array}\right.\Rightarrow 2g(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}M\frac{m^2}{M^2}v_c^2$$
$$\rightarrow 2g(h+g)=\frac{1}{2}mv_c^2+\frac{1}{2}m\left( 1+\frac{m^2}{M} \right)v_c^2$$
It isn't correct.

The error is in one of your assumptions. I suggest making sure all of your initial assumptions are dimensionally consistent.

Edit: The fact that you have a term $1 + m^2/M$ should ring large warning bells. The 1 is just a number and the term $m^2/M$ has units of mass. You cannot add a mass to a dimensionless number ... although I suspect that is arithmetic error rather than a result of your initial dimensional inconsistency.

$$F_{centripetal}=m\frac{(v_c+V_c)^2}{R}$$

Correct. Compared to a stationary rail, the bead needs to accelerate more since it will need to go up faster due to the movement of the rail.

 

[Karol]

$$\left\{ \begin{array}{l} mg(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=MV_c \end{array}\right.\Rightarrow v_c^2=\frac{4MG(h+R)}{m+M},\;V_c^2=\frac{m^2}{M^2}v_c^2$$
$$F_{centripetal}=m\frac{(v_c+V_c)^2}{R}=\frac{mv_c^2}{R}\left( 1+\frac{m}{M} \right)^2$$
$$\vec{F}t=m\vec{v}\rightarrow \vec{F}t=m(\vec{v_c}-\vec{v_b})$$
$$\sqrt{v_c^2+V_c^2}=v_c\sqrt{1+\frac{m^2}{M^2}}$$

 

[Orodruin]

I am on the bus so I cannot check with what I got but it looks reasonable.

 

[Karol]

Thanks Orodruin and the rest

 

[Karol]

I think i made a mistake, the bead and the cart's velocities are opposite:
$$\left\{ \begin{array}{l} mg(h+R)=\frac{1}{2}mv_c^2+\frac{1}{2}MV_c^2 \\ mv_c=-MV_c \end{array}\right.\Rightarrow v_c^2=\frac{4MG(h+R)}{m+M},\;V_c^2=\frac{m^2}{M^2}v_c^2$$
The Centripetal force:
$$F_{centripetal}=m\frac{(v_c+V_c)^2}{R}=\frac{mv_c^2}{R}\left( 1+\frac{m(m-2)}{M} \right)$$
Which is dimensionally wrong
Edit: v_c and V_c are absolute, so the relative velocity between the bead and the cart is $v_c-V_c$ so the original post is fine, since even with $V_c=-\frac{m}{M}v_c$ it comes out the same