Question About Momentum: Ball Rolling up a Curved Ramp

[chaneth8]

Homework Statement

Question About Momentum

Relevant Equations

$\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ = $\Delta$ momentum

This isn't exactly a homework question, but it was inspired by a homework problem. It's my first time here, so if there's a better place for this question feel free to point it out.

Let's consider a ball rolling up the following circular slope. Let's assume it rolls up the slope until it reaches the top - afterwards, it rolls back down. Let's consider the change in momentum of the ball in the direction of j.

At the time $t_\rm{initial}$ , when the ball is at the foot of the slope, the ball has velocity 0 in the direction of j - all its velocity is directed in the horizontal direction. At the time $t_\rm{final}$, when the ball is at the top of the slope, the ball has velocity 0 in the direction of j too - this is because we're at the point just before the ball rolls back down. This means that the change in momentum between these 2 points in time is 0.

However, we also know that the change in momentum between 2 points in time can be given by integrating the external forces acting on the ball in the j-direction. There is always an external force acting on the ball in the direction of j as it climbs up the slope - we've got gravity and the normal force, and the j-components of those 2 don't always cancel out. So the integral $\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ is non-zero

How can this be? We've know in general that $\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ = $\Delta$ momentum but we've shown that in the j-direction,$\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ is nonzero, while $\Delta$ momentum is equal to 0.

 

[Aurelius120]

There is always an external force acting on the ball in the direction of j as it climbs up the slope - we've got gravity and the normal force, and the j-components of those 2 don't always cancel out. So the integral $\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ is non-zero

$F_{external}=0$ isn't the only condition for the integral to be zero.
Consider
$$\int_0^{\pi}\cos x dx$$

 

[chaneth8]

$F_{external}=0$ isn't the only condition for the integral to be zero.
Consider
$$\int_0^{\pi}\cos x dx$$

That makes sense! One last question though - why would the integral $\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ evaluate to 0?

I'm asking this because if we consider the direction of j to be positive, then it seems to me that the value of $F_\rm{external}$ will always be negative or zero. We've got the force of gravity pointing downwards at every point in time, and the j-component of the normal force pointing upwards at every point in time. However, the j-component of the normal force can never exceed the magnitude of the gravity force - so what's causing the value of the integral to cancel out and become zero?

 

[Aurelius120]

Ball stops moving at topmost point as all kinetic energy is converted to potential energy.
Since the ball only had $i$ component of velocity initially, net change in momentum is along $i$ while change in $j$ component is zero. Therefore that integral has to be zero

 

[chaneth8]

I understand that, but I'm considering this from the perspective of the external forces acting on the ball in the j direction. To my understanding, at every point in time, the net force in the j-direction has a negative or zero value, assuming we consider the direction of the j as positive. So I don't understand how integrating the external forces in the j direction will give an integral that evaluates to zero. How can the integral evaluate to zero, if there isn't at least some time interval where the external force is positive?

 

[gmax137]

I think you need to write out the vx and vy equations and actually do the integrals to see where your intuition is misleading you.

 

[jbriggs444]

However, the j-component of the normal force can never exceed the magnitude of the gravity force

What makes you say this?

Unrelated to this, for rolling without slipping, the force of static friction should not be ignored. Something must act on the rolling ball to cause its rotation to cease and then to reverse.

 

[chaneth8]

What makes you say this?

This is my reasoning:

At any point in time, let us denote the normal force acting on the ball as $F_\rm{N}$, and the gravity force as $F_\rm{g}$. If we take the angle between the vectors for $F_\rm{N}$ and $F_\rm{g}$, we can see that the magnitude of $F_\rm{N}$ = $mg cos(\theta)$.

Since the magnitude of the normal force is less than the magnitude of the gravity force, the magnitude of the j-component of the normal force should also be less than the magnitude of the j-component of the gravity force (since gravity acts exclusively in the j-direction).

Could you point out where my reasoning is wrong? I'd appreciate it.

 

[Aurelius120]

If the ball is moving on a curve, the required Centripetal force will be provided by difference in respective components of $mg$ and $N$
So depending on Radius of Curvature, Normal might exceed weight

That could be a reason but I am no expert

 

[jbriggs444]

This is my reasoning:
View attachment 347565
At any point in time, let us denote the normal force acting on the ball as $F_\rm{N}$, and the gravity force as $F_\rm{g}$. If we take the angle between the vectors for $F_\rm{N}$ and $F_\rm{g}$, we can see that the magnitude of $F_\rm{N}$ = $mg cos(\theta)$.

You need to reason more carefully.

Write down the Newton's second law force balance ($\sum F = ma$) in the vertical direction. For convenience, let $a = 0$. See if you get what you say you get.

 

[haruspex]

If the ball is moving on a curve, the required Centripetal force will be provided by difference in respective components of $mg$ and $N$
So depending on Radius of Curvature, Normal might exceed weight

That could be a reason but I am no expert

That is indeed the reason.
In the question as stated, there is also static friction, which will have a vertically upward component. But we can recast the question as being frictionless and ignore the rotation of the ball. It is still true that as the ball starts up the ramp it is accelerating upward, so N exceeds mg.

 

[gleem]

What force is causing the ball to go up the ramp?

 

[renormalize]

How can this be? We've know in general that $\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ = $\Delta$ momentum but we've shown that in the j-direction,$\int_{t_\rm{initial}}^{t_\rm{final}} F_\rm{external} \, dt$ is nonzero, while $\Delta$ momentum is equal to 0.

Let's look at this scenario in more detail by simplifying the problem to be a bead of mass $m$ sliding, without rotation, on a frictionless, circular ramp of radius $R$. Below is the configuration that I will consider:

I begin by ignoring dynamics (forces, energies and equations of motion) and focus, for the moment, on kinematics. The coordinates of the bead as a function of time $t$ are determined by the single parameter $\alpha(t)$:$$x\left(t\right)=R\sin\alpha\left(t\right),\quad y(t)=R\left(1-\cos\alpha\left(t\right)\right)\tag{1a,b}$$where $0\leq t\leq t_f$. Here $t=0$ represents the moment that the bead first enters the ramp, with initial velocity $v_0$ in the $x$-direction, and $t=t_f$ is the time when the bead reaches it's maximum height in the $y$-direction, with an instantaneous velocity of zero. The range of the angle $\alpha$ is thus $0=\alpha (0)\leq\alpha(t)\leq\alpha(t_f)$, where $\alpha(t_f)$ is to be determined from the dynamics of the problem. The components of the initial and final velocities are therefore:$$\left(\dot{x}\left(0\right),\dot{y}\left(0\right)\right)=\left(v_{0},0\right),\quad\left(\dot{x}\left(t_{f}\right),\dot{y}\left(t_{f}\right)\right)=\left(0,0\right)\tag{2a,b}$$I compare this expression to eqs.(1) and their first-derivatives:$$\dot{x}\left(t\right)=R\cos\alpha\left(t\right)\dot{\alpha}\left(t\right),\quad\dot{y}\left(t\right)=R\sin\alpha\left(t\right)\dot{\alpha}\left(t\right)\tag{3a,b}$$to conclude that $\dot{\alpha}$ satisfies the conditions: $$\dot{\alpha}\left(0\right)=\frac{v_{0}}{R},\quad\dot{\alpha}\left(t_{f}\right)=0\tag{4a,b}$$Now I examine accelerations by differentiating eqs.(3) to find:$$\ddot{x}\left(t\right)=-R\sin\alpha\left(t\right)\dot{\alpha}^{2}\left(t\right)+R\cos\alpha\left(t\right)\ddot{\alpha}\left(t\right),\quad\ddot{y}\left(t\right)=R\cos\alpha\left(t\right)\dot{\alpha}^{2}\left(t\right)+R\sin\alpha\left(t\right)\ddot{\alpha}\left(t\right)\tag{5a,b}$$Note that at $t=0$ the $y$-acceleration is $\ddot{y}\left(0\right)=v_{0}^{2}/R$; i.e., the bead experiences an instantaneous centripetal acceleration in the vertical direction, as expected. But in the horizontal direction, $\ddot{x}\left(0\right)=R\,\ddot{\alpha}\left(0\right)$, which demands that $\ddot{\alpha}(0)=0$ since there is no force exerted on the bead in the $x$-direction at the moment it enters the ramp. Overall then, the kinematics of the sliding bead dictate the following conditions on the angle $\alpha(t)$:$$\alpha\left(0\right)=0,\quad\dot{\alpha}\left(0\right)=\frac{v_{0}}{R},\quad\ddot{\alpha}\left(0\right)=0,\quad\dot{\alpha}\left(t_{f}\right)=0\tag{6a,b,c,d}$$To progress further requires dynamics. The equation of motion for $\alpha(t)$ can only be solved implicitly in terms of an elliptic integral of the first kind. But there's no need to go that far to address the OP's concern. Instead, I use the conservation (constancy) of the total energy $E\equiv E_{\text{kinetic}}\left(t\right)+E_{\text{potential}}\left(t\right)$ to write:$$E=\frac{1}{2}m\left(\dot{x}^{2}\left(t\right)+\dot{y}^{2}\left(t\right)\right)+mg\,y\left(t\right)=E\left(0\right)=\frac{1}{2}mv_{0}^{2}\tag{7}$$Dividing by $g$ and using eqs.(3) to eliminate $\dot{x},\dot{y}$ in favor of $\dot{\alpha}$ converts eq.(7) to:$$\frac{E}{g}=\frac{mv_{0}^{2}}{2g}=m\,y\left(t\right)+\frac{mR^{2}\dot{\alpha}^{2}\left(t\right)}{2g}\tag{8}$$Differentiating this twice w.r.t. time yields the expression for the the force in the $y$-direction $F_y$:$$m\,\ddot{y}\left(t\right)\equiv F_{y}\left(t\right)=-\frac{mR^{2}\left(\ddot{\alpha}^{2}\left(t\right)+\dot{\alpha}\left(t\right)\dddot{\alpha}\left(t\right)\right)}{g}=-\frac{d}{dt}\left[\frac{mR^{2}\dot{\alpha}\left(t\right)\ddot{\alpha}\left(t\right)}{g}\right]\tag{9}$$Eq.(9) gets to the crux of the OP's question. OP asserts that the time-integral of the integrand on the right side of (9) must be non-zero. But that assertion is false because the integrand is not manifestly of a single sign. Instead, on the solution set for $\alpha(t)$, the integrand varies in sign such that the time-integral of the force from $0$ to $t_f$ does indeed exactly vanish:$$mv_{y}\left(t_{f}\right)-mv_{y}\left(0\right)\equiv\intop_{0}^{t_{f}}F_{y}\left(t\right)dt=\frac{mR^{2}}{g}\left[\dot{\alpha}\left(0\right)\ddot{\alpha}\left(0\right)-\dot{\alpha}\left(t_{f}\right)\ddot{\alpha}\left(t_{f}\right)\right]=0\tag{10}$$ in view of the conditions (6c,d).

(Well that kept me up past my bedtime.)