Pinned massless stick with a concentrated mass at the other end.

[cshum00]

Homework Statement

A massless stick is pinned at point A. And it has a concentrated ball mass of 1kg attached to it at point B.

Given the initial values:
Initial angular position θ=45◦
Initial angular velocity ω=0
Initial angular acceleration α=0

Show that because of the setup of the problem, the concentrated mass B follows a circular path of:
x = \sqrt{2}cos\thetay = \sqrt{2}sin\theta

Homework Equations

The tangential acceleration is induced by the gravitational force on the mass:
ma_t = mgcos\theta

The stick reacts to the compressive force induced by the ball mass, which is also the centripetal acceleration:
ma_c = AB = mgsin\theta

The Attempt at a Solution

Find the forces acting at point B:
\sum F_x= ma_xAB_x = ma_xABcos\theta = ma_xmgsin\theta cos\theta = ma_xa_x = gsin\theta cos\theta

\sum F_y= ma_yAB_y - mg = -ma_y-ABsin\theta + mg = ma_ym(g - gsin^2\theta) = ma_ya_y = g(1 - sin^2\theta)a_y = gcos^2\theta

To find the x and y positions simply double integrate the respective accelerations a_x and a_y with respect to time.
x = \int \int a_x = \int \int gsin\theta cos\theta dt^2y = \int \int a_y = \int \int gcos^2\theta dt^2

However, the problem is that θ is a function of time or θ(t). In addition, although there is a constant vertical gravitational acceleration, the angular acceleration is not constant. Therefore, the equation for constant angular acceleration is not applicable:
\theta = \frac{1}{r}(\theta _0 + \omega _0 t + \frac{1}{2}\alpha _0 t^2)

So, in order to find θ, i have to use the tangential acceleration relation with the angular acceleration:
r\alpha = a_t\alpha = \frac{a_t}{r}r\theta = \int \int \frac{a_t}{r}dt^2

However, because the tangential acceleration is known in terms of θ, it sets me back to my original problem that i am trying to integrate θ with respect to time:
\theta = \frac{1}{r^2}\int \int gcos\theta dt^2

 

[mfb]

If you posted the full problem statement, you don't have to find the time-evolution (= θ(t)) at all.

To get θ(t) (if you really need it), you can find a differential equation and solve this. Alternatively, use energy conservation to find $\dot{θ}(θ)$ and solve this backwards to get t(θ) and then θ(t).

 

[cshum00]

@mfb
When you are talking about the energy equations, i believe you are referring to:\Delta KE = 0\frac{1}{2}I\omega _f^2 - \frac{1}{2}I\omega _i^2= 0\frac{1}{2}I(\omega _f^2 - \omega _i^2)= 0

However, the problem is finding the angular velocity:\omega = \int \alpha dt\omega = \int \frac{1}{r}gcos\theta dt

Which i am back to round one. Maybe there is a way to go from position, velocity and acceleration without going through integration/differentiation with respect to time; which i don't know. Can you please show me your method?

 

[mfb]

@mfb
When you are talking about the energy equations, i believe you are referring to:\Delta KE = 0

That is wrong, there is a second form of energy involved in your setup.

However, the problem is finding the angular velocity:\omega = \int \alpha dt\omega = \int \frac{1}{r}gcos\theta dt

There is an easier way to find the angular velocity for a given angle, you don't need to integrate anything.

Maybe there is a way to go from position, velocity and acceleration without going through integration/differentiation with respect to time; which i don't know. Can you please show me your method?

Energy conservation.


Please post the full problem statement. It's still unclear what exactly you are supposed to do.

 

[cshum00]

@mfb
That is the full problem statement because I made it up. I want to mathematically show that the path taken by the ball mass at the end of the pinned stick is:x=\sqrt{2}cos\thetay=\sqrt{2}sin\thetaIs there any unknown or missing variable you want to add to the problem?

It is true that i can find the angular velocity of a given angle, but i want to be able to setup the equations for any angle. As for the energy equation, are you referring to this one?\Delta mgh = \Delta \frac{1}{2}mv^2 + \Delta \frac{1}{2}I \omega^2If not, please just tell me the equation.

 

[TSny]

@mfb
That is the full problem statement because I made it up. I want to mathematically show that the path taken by the ball mass at the end of the pinned stick is:x=\sqrt{2}cos\thetay=\sqrt{2}sin\thetaIs there any unknown or missing variable you want to add to the problem?

From your original picture, use trigonometry to find the $x$ coordintate of the ball in terms of $l$ and $\theta$.

Similarly for the $y$ coordinate.

 

[cshum00]

Ok. Thanks to your reply TSny, i think i know what mfb meant by the problem statement being incomplete.

So, because of the problem setup we can agree that the motion path of the ball mass will be of circular motion of the form:x=rcos\thetay=rsin\thetaWhere θ is a function of time t.So, if the problem were to be of constant speed, the motion path would be:x=rcos(\theta_0+\omega_0 t)y=rsin(\theta_0+\omega_0 t)

If, the problem were to be of constant acceleration, the motion path would be:x=rcos(\theta_0+\omega_0 t+\frac{1}{2}\alpha_0 t^2)y=rsin(\theta_0+\omega_0 t+\frac{1}{2}\alpha_0 t^2)

So, the question for this problem should be, find exact form of the motion path. In order to do so, i have to find what is θ(t) first. I will update the original post as well.

------------------------------------------------------------------------------------------

In addition, i believe that my previous energy equations are wrong. They should be:\Delta PE + \Delta KE = 0Therefore, mg(h_f - h_i) + \frac{1}{2}I(\omega_f ^2 - \omega_i ^2) = 0I had the kinetic energy repeated twice in my previous equation:\frac{1}{2}mv^2 = \frac{1}{2}mr^2 \omega^2 = \frac{1}{2}I \omega^2

Edit: How do i update the original post?

 

[TSny]

I believe it's too late to edit your original post. But that's ok.

Your energy equation looks good. Note that you can let h = y, which can be expressed in terms of θ.

To find θ(t) you can try mfb's suggestion in post #2. However, I believe that it will not be possible to express θ(t) in terms of "elementary" functions.

 

[mfb]

It is true that i can find the angular velocity of a given angle, but i want to be able to setup the equations for any angle.

That is the same thing, as you can keep the "given angle" as variable.

So, the question for this problem should be, find exact form of the motion path. In order to do so, i have to find what is θ(t) first.

That is the last relevant step (and it is a step you never mentioned before as goal). Once you have θ(t), you are done.

mg(h_f - h_i) + \frac{1}{2}I(\omega_f ^2 - \omega_i ^2) = 0

Looks good.

To find θ(t) you can try mfb's suggestion in post #2. However, I believe that it will not be possible to express θ(t) in terms of "elementary" functions.

As this is not a "real" homework question, I have the same feeling. I had a similar problem in an exam long ago, and we were supposed to (and had to) make approximations to get an answer. It was not the same problem, but it was so similar I would be surprised if that makes a difference in terms of analytic solutions.

 

[cshum00]

I have tried various things and they don't seem to come out quite right. Maybe you guys can help me spot any mistakes.\alpha=\frac{a_t}{r}=\frac{gcos\theta}{r}
If:x=rcos\thetaa_x=gcos\theta sin\thetaSo:rcos\theta=\int\int gcos\theta\sin\theta dt^2\frac{d^2}{dt^2}(rcos\theta)=gcos\theta\sin\theta\frac{d}{dt}(-rsin\theta\frac{d\theta}{dt})=gcos\theta\sin\theta-rcos\theta(\frac{d\theta}{dt})^2-rsin\theta\frac{d^2\theta}{dt^2}=gcos\theta\sin\theta-rcos\theta\omega^2-rsin\theta\alpha=gcos\theta\sin\theta-rcos\theta\omega^2=gcos\theta\sin\theta+rsin\theta(\frac{gcos\theta}{r})-rcos\theta\omega^2=2gcos\theta\sin\theta\omega^2=\frac{-2gsin\theta}{r}
But if i try the same with:y=rsin\thetaa_y=gcos^2\thetaI get:
rsin\theta=\int\int gcos^2\theta dt^2\frac{d^2}{dt^2}(rsin\theta)=gcos^2\theta\frac{d}{dt}(rcos\theta\frac{d\theta}{dt})=gcos^2\theta-rsin\theta(\frac{d\theta}{dt})^2+rcos\theta\frac{d^2\theta}{dt^2}=gcos^2\theta-rsin\theta\omega^2+rcos\theta\alpha=gcos^2\theta-rsin\theta\omega^2=gcos^2\theta-rcos\theta(\frac{gcos\theta}{r})-rsin\theta\omega^2=0\omega^2=0

I don't get the same answer. So, there is a possibility i left a minus sign somewhere. So let's assume this instead:-rsin\theta\omega^2=gcos^2\theta+rcos\theta(\frac{gcos\theta}{r})-rsin\theta\omega^2=2gcos^2\theta\omega^2=\frac{2gcos^2\theta}{-rsin\theta}\frac{sin\theta}{sin\theta}\omega^2=\frac{-2gsin\theta}{r}cot^2\thetaWhich is still different by cot²θ.

 

[TSny]

I have tried various things and they don't seem to come out quite right. Maybe you guys can help me spot any mistakes.\alpha=\frac{a_t}{r}=\frac{gcos\theta}{r}
If:x=rcos\thetaa_x=gcos\theta sin\theta

You have assumed that a_x comes entirely from the x-component of the tangential acceleration a_t. But the ball also has centripetal acceleration that will contribute an additional part to a_x. If you include that and go through a similar calculation to what you did, I think you will just end up with an identity: 0 = 0.

 

[cshum00]

Actually, if you review my opening post, a_x comes from the x-component of the centripetal acceleration a_c. The centripetal force AB is the mass times the centripetal acceleration a_c.

The reason why i put the angular acceleration relationship with the tangent acceleration a_t is because it is later on useful to substitute:\frac{d^2\theta}{dt^2}=\alpha=\frac{gcos\theta}{r}

On the other hand, the y-component acceleration is derived by the centripetal reaction plus the gravitational force acting on the mass.

Also, using the energy equation i get similar results, unless i did something wrong:mg\Delta h+\frac{1}{2}I\Delta\omega^2=0\Delta\omega^2=\frac{-2mg\Delta h}{I}\omega^2=\frac{-2mgrsin\theta}{mr^2}\omega^2=\frac{-2gsin\theta}{r}
So I think there is something wrong on the y and a_y side instead.

Edit: The magnitude of the resultant x and y accelerations is equals to the tangential acceleration:\sqrt{a_x^2 + a_y^2}=a_t\sqrt{g^2cos^2\theta sin^2\theta + g^2cos^4\theta}=gcos\theta\sqrt{g^2cos^2\theta(sin^2\theta + cos^2\theta)}=gcos\theta\sqrt{g^2cos^2\theta(1)}=gcos\thetagcos=gcos\theta

 

[TSny]

Actually, if you review my opening post, a_x comes from the x-component of the centripetal acceleration a_c. The centripetal force AB is the mass times the centripetal acceleration a_c.

Oh, I misinterpreted how you got $a_x$. I thought you took the x-component of $a_t$ which would be $g\cos\theta\cdot\sin\theta$ which is the same as the expression you wrote for $a_x$.

Looking back at the original post, I don’t understand your analysis of the centripetal acceleration, $a_c$. There are two separate forces acting on the ball that produce the centripetal acceleration: (1) the centripetal component of the force of gravity (acting toward the center) and (2) the force that the stick exerts on the ball (which could act away from the center or toward the center depending on where the ball is located in its fall.)

Force (1) is $mg\sin\theta$ and force (2) I'll write as $F_{\rm stick}$.

Newton’s second law for the centripetal direction is $\sum F_c = ma_c$.

This gives $mg\sin\theta + F_{\rm stick} = ma_c$, where $F_{\rm stick}$ is positive if it acts toward the center and negative if it acts away from the center.

So, it is not true that $ma_c = mg\sin\theta$ (except at the one point during the fall where $F_{\rm stick}=0$).

 

[cshum00]

Newton’s second law for the centripetal direction is $\sum F_c = ma_c$.

By doing it your way, if you look only at the centripetal forces, the sum of all the centripetal forces must equal to zero. That is because the resultant does not move into or away from the distance of the stick. So using your formula:

$mgsin\theta + F_{stick} = 0$
$F_{rm stick} = -mgsin\theta$

In other words, the sticks reacts on a opposite force to the centripetal component of the gravitational force of the mass.

The way i originally did it was to project the gravitational acceleration of the mass onto a centripetal and tangent component. That is because the mass causes both the centripetal and tangential force. Without the mass, the massless stick would not have any tangential or centripetal force. If the mass were to be alone by itself, it would freefall. The combination of both creates this mass falling on a circular motion kind of fashion.

 

[TSny]

If the sum of the centripetal forces equals zero, how could there be a nonzero centripetal acceleration?

 

[cshum00]

I guess, zero tangential acceleration would be a non-zero centripetal acceleration. I mean, i know it is weird that there is no centripetal acceleration and i am with you.

Edit: Maybe we should a similar example with a pendulum. The only difference is that there is going to be tension instead of compression for the massless string.

The sum of the centripetal force using Newton's second law.

$\sum F_c = ma_c$
$mgcos\theta + T_{string} = 0$
$T_{string} = -mgcos\theta$
$ma_c = -mgcos\theta$
$a_c = -gcos\theta$

 

[TSny]

Yes, it's similar to a pendulum.

You know from kinematics that centripetal acceleration for circular motion is related to the speed and radius: $a_c = \frac{v^2}{r}$. So, there will be centripetal acceleration as long as the ball has some speed.

 

[cshum00]

Yes but the centripetal acceleration of $a_c = \frac{v^2}{r}$ is for constant tangential velocity and no tangential acceleration. In the case of the pendulum, there is tangential acceleration.

 

[TSny]

Yes but the centripetal acceleration of $a_c = \frac{v^2}{r}$ is for constant tangential velocity and no tangential acceleration. In the case of the pendulum, there is tangential acceleration.

$a_c = \frac{v^2}{r}$ is valid even if there is tangential acceleration.

 

[mfb]

Also, using the energy equation i get similar results, unless i did something wrong:mg\Delta h+\frac{1}{2}I\Delta\omega^2=0\Delta\omega^2=\frac{-2mg\Delta h}{I}\omega^2=\frac{-2mgrsin\theta}{mr^2}

Do you assume that the mass starts at the highest point here?I don't see the relevance of centripetal acceleration, and the thread begins to get messy with all those calculation fragments.

For a clean way:

Can you find a differential equation for the angular acceleration as function of the angle? This is identical to an inclined plane, as locally the mass is like on such a plane.

Alternatively, write the angular velocity as function of the angle.

 

[cshum00]

$a_c = \frac{v^2}{r}$ is valid even if there is tangential acceleration.

I am not sure about that. Let's say we have a scenario that there is no gravity and that there is constant angular acceleration. How do you prove that the centripetal acceleration is: $a_c = \frac{v^2}{r}$? For example, let's say that the motion path formula is:
$x = rcos(\theta_0 + \omega_0 t + \frac{1}{2}\alpha_0 t^2)$
$y = rsin(\theta_0 + \omega_0 t + \frac{1}{2}\alpha_0 t^2)$
Or what would be the magnitude of the centripetal acceleration in your calculations for the pendulum case?

Do you assume that the mass starts at the highest point here?

I assume that it starts at an arbitrary point as long as it is not 90°. If you want an easy case for numerical calculations, then i would say 45°.

I don't see the relevance of centripetal acceleration, and the thread begins to get messy with all those calculation fragments.

I need the centripetal acceleration to find he reaction of the stick to the ball mass. Sorry for it being messy for you. I wanted to brainstorm as many expressions related to the problem as possible and that way pick the ones most useful in solving the problem. It has always been helpful to me.

Can you find a differential equation for the angular acceleration as function of the angle? This is identical to an inclined plane, as locally the mass is like on such a plane.

I actually can find the angular acceleration as a function of the angle and no differential equation.
$r\alpha=a_t$
$\alpha=\frac{a_t}{r}$
$\alpha=\frac{gcos\theta}{r}$
But that is not what i want. I want to get x and y as a function of time.

Edit: @mfb Were you referring the opening thread to be messy or the one with various differential equations being confusing?

 

[mfb]

How do you prove that the centripetal acceleration is: $a_c = \frac{v^2}{r}$?

Define an arbitrary path with x(t) and y(t). The distance to the origin stays the same, so $x(t)^2 + y(t)^2 = r^2$ with some fixed r, for all t.

Calculating the second derivative of this expression gives
$$2\dot{x}^2 + 2x\ddot{x} + 2\dot{y}^2 + 2y\ddot{y} = 0$$
where dots are derivatives with respect to time.

As $v^2=\dot{x}^2 + \dot{y}^2$, we can re-write this equation as
$$v^2 + x\ddot{x} +y\ddot{y} = 0$$
or
$$x\ddot{x} +y\ddot{y} = -v^2$$

We can define a normalized radial vector:
$$\frac{1}{r}\left(\begin{array}{l}
x\\
y\end{array}\right)$$
The acceleration in radial direction (and therefore the centripetal acceleration) is then just the scalar product of the acceleration with this vector:
$$a_c = \frac{1}{r} \left(\begin{array}{l}
x\\
y\end{array}\right) \cdot
\left(\begin{array}{l}
\ddot{x}\\
\ddot{y}\end{array}\right) = \frac{1}{r} (x \ddot{x} + y \ddot {y})$$
Using the equation from above, $$a_c = - \frac{v^2}{r}$$
The negative sign indicates an acceleration towards smaller radii, which is inwards.

I assume that it starts at an arbitrary point as long as it is not 90°.

Then the last quoted step is wrong, that's the reason I asked.

I need the centripetal acceleration to find he reaction of the stick to the ball mass.

You don't need this reaction.

But that is not what i want. I want to get x and y as a function of time.

It is significantly easier if you start with the angle, and convert this to x- and y-coordinates afterwards. If you want to use x and y everywhere, fine, but then I'll ignore this thread as I consider it a huge waste of time and a big source of potential errors.

I actually can find the angular acceleration as a function of the angle and no differential equation.
$r\alpha=a_t$
$\alpha=\frac{a_t}{r}$
$\alpha=\frac{gcos\theta}{r}$

That is a differential equation, it contains both θ and its second derivative, $\alpha$.

Edit: @mfb Were you referring the opening thread to be messy or the one with various differential equations being confusing?

The posts after the first one.

 

[cshum00]

We can define a normalized radial vector:
$$\frac{1}{r}\left(\begin{array}{l}
x\\
y\end{array}\right)$$

I believe you are referring to column vector in your expression and not combination. I like your proof and everything looks right over there. How do i get the tangential acceleration using your method?

It is significantly easier if you start with the angle, and convert this to x- and y-coordinates afterwards. If you want to use x and y everywhere, fine, but then I'll ignore this thread as I consider it a huge waste of time and a big source of potential errors.

I have been trying it both ways using rectangular and polar coordinates. The problem with polar coordinate is that i get a function such that θ(θ). From there, i have no idea how to convert back to rectangular coordinates. Also, not sure about the thread being a huge waste of time; but it is highly possible that there are errors here and there.

That is a differential equation, it contains both θ and its second derivative, $\alpha$.

If you look at it at that way, yes it is.

Edit: So, was my angular speed result from my energy equation wrong?

Edit2:

The posts after the first one.

Well, TSny and I were discussing about how i got the x and y force components after the first post. That is because the first post i used the centripetal acceleration to find the x and y forces.