How to show that P moves with a constant speed u

[Nicci]

Homework Statement

A particle P of mass m slides on a smooth horizontal table. P is connected to a second particle Q of mass M by a light inextensible string which passes through a small hole O in the table, so that Q hangs below the table while P moves on top. Investigate motions of this system in which Q remains at rest vertically below, while P describes a circle with center O and radius b. Show that P moves with constant speed u, $$u^2 = (Mgb)/m$$

Relevant Equations

For particle Q: T - Mg = 0
Thus, T = Mg

For particle Q:
The resultant force on particle Q would be zero since it is at rest. Thus$$ T - Mg = 0$$ which gives $$T = Mg$$
For particle P:
This is where I am struggling. I can't seem to write out the polar equations of motion. I have to show that $$u^2 = (Mgb)/m$$
I know that $$\vec a = (\ddot r-r(\dotΘ)^2) \hat r + (r\ddotΘ + 2\dot r\dotΘ)\hatΘ$$
There has to be two polar equations of motion for P. Should I assume that $$\ddot r= 0$$ and $$\dot r = 0$$ or am I missing something? Then when I say that r=b I will get $$\vec a = (0-b(\dotΘ)^2) \hat r+0$$
I am not sure if I am even on the right track. Can someone please provide me with a hint or some help?
Thank you very much.

 

[scottdave]

I think you may find the following video interesting. They do work out the mathematics behind a spinning mass on a string.

 

[TSny]

Your approach is fine.

Should I assume that $$\ddot r= 0$$ and $$\dot r = 0$$ or am I missing something?

This is correct. The problem states that particle P moves in a circle. Therefore, $r$ remains constant. So, $\dot r = \ddot r = 0$.

Then when I say that r=b I will get $$\vec a = (0-b(\dotΘ)^2) \hat r+0$$

What is the reason for asserting that the $\hat \Theta$ component of the acceleration is zero?

 

[Nicci]

What is the reason for asserting that the $\hat \Theta$ component of the acceleration is zero?
[/QUOTE]

I see I made a mistake there. The equation should then be $$\vec a = -b(\dot Θ)^2 \hat r + b\ddot Θ\hat Θ$$
The second equation is $$\vec v = \dot r \hat r + r\dot Θ\hat Θ$$ which gives $$\vec v = b\dot Θ\hatΘ$$
The circumferential speed is $$u=b\dotΘ$$
How do I relate the T=Mg with the velocity and acceleration equations? Would the tension be the same for Particle P and Particle Q as the string is not extending?
Thank you very much.

 

[wrobel]

The system is governed by two equations. Let $r,\varphi$ be polar coordinates of the particle $P$ on the table. Equation of angular momentum conservation is
$$r^2\dot\varphi=const_1.$$ The energy is also conserved:
$$\frac{m}{2}\Big((r\dot\varphi)^2+\dot r^2\Big)+\frac{M}{2}\dot r^2+rMg=const_2.$$
Express $\dot\varphi$ from the first equation and substitute it to the second one to obtain effective potential.

 

[TSny]

I see I made a mistake there. The equation should then be $$\vec a = -b(\dot Θ)^2 \hat r + b\ddot Θ\hat Θ$$
The second equation is $$\vec v = \dot r \hat r + r\dot Θ\hat Θ$$ which gives $$\vec v = b\dot Θ\hatΘ$$
The circumferential speed is $$u=b\dotΘ$$

OK. But part of what you are asked to show is that the speed $u$ is constant. So, you need to find an argument for why $\dotΘ$ is constant.

How do I relate the T=Mg with the velocity and acceleration equations? Would the tension be the same for Particle P and Particle Q as the string is not extending?

Yes, the tension is the same at both ends of the string.

Have you drawn a good free-body diagram for particle P, showing all forces acting on P? If so, then you are ready to set up Newton's second law for the $\hat r$ and $\hat \Theta$ directions. This might help with showing that the speed is constant.

 

[Nicci]

OK. But part of what you are asked to show is that the speed $u$ is constant. So, you need to find an argument for why $\dotΘ$ is constant.

Yes, the tension is the same at both ends of the string.

Have you drawn a good free-body diagram for particle P, showing all forces acting on P? If so, then you are ready to set up Newton's second law for the $\hat r$ and $\hat \Theta$ directions. This might help with showing that the speed is constant.

I have tried my best the last few days. I did manage to draw a free body diagram. If I am correct then $$\vec F = -mg$$
Doesn't that mean that $$\vec F = m\vec a$$ and that gives $$-mg = m(-b\dotΘ^2\hat r + b\ddot Θ \hatΘ)$$
Somewhere I am missing something because I have to somehow "get rid of" the $$ \hat r$$ and $$\hat Θ$$
Thank you very much.

 

[TSny]

I did manage to draw a free body diagram.

Can you show or describe your free body diagram for particle P? The diagram should show three forces acting on P. Two of these forces act perpendicular to the table. What two forces are these? Consider a z-axis perpendicular to the table with $\hat z$ a unit vector pointing upward along this axis. So, these two forces can be expressed in terms of $\hat z$.

What is the third force acting on P? How would you express it in terms of $\hat r$ or $\hat \Theta$?

Once you know the forces that act along each unit vector direction, you can set up Newton's second law for each of the three directions. For example, for the $\hat z$ direction: $F^{net}_z = \sum F_z = m a_z$. Thus, you should obtain 3 separate equations, one for each direction $\hat z$, $\hat r$, $\hat \Theta$.

 

[Nicci]

Can you show or describe your free body diagram for particle P? The diagram should show three forces acting on P. Two of these forces act perpendicular to the table. What two forces are these? Consider a z-axis perpendicular to the table with $\hat z$ a unit vector pointing upward along this axis. So, these two forces can be expressed in terms of $\hat z$.

What is the third force acting on P? How would you express it in terms of $\hat r$ or $\hat \Theta$?

Once you know the forces that act along each unit vector direction, you can set up Newton's second law for each of the three directions. For example, for the $\hat z$ direction: $F^{net}_z = \sum F_z = m a_z$. Thus, you should obtain 3 separate equations, one for each direction $\hat z$, $\hat r$, $\hat \Theta$.

This is the free-diagram and the picture that I have drawn. I see did not include the i and j axis on the picture.
The two forces that are perpendicular to the table are the $$\vec F = -mg$$ and the unit vector $$\hat k$$
I think the third force would be the tension, T, on the string? Would the tension not then be T = -mgsinΘ ?
Would the equation be $$F(net) = \hat k -T - mg$$
Then $$F(net) = \hat k +mgsinΘ -mg$$
I'm sorry, I think I am way off now.

 

[TSny]

This is the free-diagram and the picture that I have drawn. I see did not include the i and j axis on the picture.
The two forces that are perpendicular to the table are the $$\vec F = -mg$$ and the unit vector $$\hat k$$

Your diagram is OK except that $\hat k$ is not a force, it is just a unit vector pointing perpendicular to the table. However, there is a force acting upward on P in the direction of $\hat k$. Can you state the name of this force?

$\hat k$ is the same unit vector as $\hat z$ that I used in my previous post. Feel free to use either notation. I will switch to using $\hat k$.

When you write $\vec F = -mg$, you can see something wrong here. The left side is a vector, but the right side is just a number. If the left side of an equation is a vector quantity, then the right side must also be a vector quantity. The correct way to write the gravitational force acting on P is $\vec F_{grav} = -mg \hat k$. The unit vector is a necessary part of the right hand side in order for the right hand side to have both a magnitude and a direction.

Likewise, when you identify the force that acts upward on P in the $\hat k$ direction, you can write it in terms of the unit vector $\hat k$.

I think the third force would be the tension, T, on the string?

Yes.

Would the tension not then be T = -mgsinΘ ?

No. You already know the magnitude of $T$ from your consideration of particle Q. Can you use one of the unit vectors $\hat k$, $\hat r$, or $\hat \Theta$ to indicate the direction of the tension force?

Would the equation be $$F(net) = \hat k -T - mg$$

No. This equation doesn't make sense for several reasons. $\hat k$ is not a force, so the first term on the right cannot be correct. Also, you can't add a vector quantity, such as $\hat k$, to a scalar quantity such as T. In addition, the tension force and the gravitational force do not act along the same direction. So, they cannot be added together as scalar quantities.

 

[Nicci]

Your diagram is OK except that $\hat k$ is not a force, it is just a unit vector pointing perpendicular to the table. However, there is a force acting upward on P in the direction of $\hat k$. Can you state the name of this force?

It should be the normal force. $$\vec F_N = mg\hat k$$ I think it should be positive since it is a force acting upward.

$\vec F_{grav} = -mg \hat k$.

[/QUOTE]
Yes.
No. You already know the magnitude of $T$ from your consideration of particle Q. Can you use one of the unit vectors $\hat k$, $\hat r$, or $\hat \Theta$ to indicate the direction of the tension force?
[/QUOTE]

The tension is along the $$\vec r$$
I'm not sure how to write it so that it shows the direction. Would it be similar to the gravitational force, but only in the direction of $$\vec r$$
and should the angle Θ not be used?

 

[TSny]

It should be the normal force. $$\vec F_N = mg\hat k$$ I think it should be positive since it is a force acting upward.

Yes, it's the normal force, $\vec F_N = F_N \hat k$. But how do you know that the magnitude of the normal force is equal to $mg$? It is worth going through, in a logical manner, how to deduce the magnitude of the normal force. You might think I'm being unnecessarily pedantic. But you can use very similar reasoning when working with the $\hat r$ and $\hat \Theta$ directions. We apply Newton's second law along the z-direction (i.e., the $\hat k$ direction): $\sum F_z = m a_z$. From the free-body diagram, we see that $\sum F_z = F_N - mg$. So, Newton's second law for the z-direction is $$ F_N - mg = ma_z$$.

We are told that particle P moves in circular motion on the table. Therefore, it never changes its vertical position. Thus we conclude that $a_z = 0$. So the second law equation reduces to $F_N - mg = 0$. Thus, we have deduced that $F_N = mg$.

Similar analysis for the $\hat \Theta$ direction will allow you to show that P must move at constant speed. From analysis of the $\hat r$ direction you can deduce the value of the speed.

The tension is along the $$\vec r$$

This is not quite correct. The unit vector $\hat r$ is defined as pointing "outward", away from the origin (the hole in the table). But, as you correctly show in your diagram, the tension force on P points "inward" toward the hole. So, the direction of $\vec T$ is $-\hat r$. So, the tension force is $\vec T = -T \hat r$.

I'm not sure how to write it so that it shows the direction.

If you want to denote a unit vector in Latex, use \hat in place of \vec.

and should the angle Θ not be used?

The angle $\Theta$ does not enter into the writing of the tension force since the tension always points in the $-\hat r$ direction.

OK. See if you can set up Newton's second law for the $\hat \Theta$ direction and try to use it to show that P has a constant speed.

After that, you can set up Newton's second law for the $\hat r$ direction.

 

[Nicci]

[/QUOTE]
OK. See if you can set up Newton's second law for the $\hat \Theta$ direction and try to use it to show that P has a constant speed.

After that, you can set up Newton's second law for the $\hat r$ direction.
[/QUOTE]

For the $\hat \Theta$ direction, shouldn't it be the $\vec v = b\dot Θ^2\hat Θ$ that I wrote earlier for the velocity?

 

[TSny]

For the $\hat \Theta$ direction, shouldn't it be the $\vec v = b\dot Θ^2\hat Θ$ that I wrote earlier for the velocity?

The $\dot \Theta$ should not be squared here. But even when you correct this, it does not show that the speed is constant. You need to show that $\dot \Theta$ is a constant. To do that, set up Newton's second law for the $\hat \Theta$ direction: $$\sum F_\theta = m a_{\theta}$$
You can make use of the general expression for $a_{\theta}$ that you posted in your first post.

 

[Nicci]

The $\dot \Theta$ should not be squared here. But even when you correct this, it does not show that the speed is constant. You need to show that $\dot \Theta$ is a constant. To do that, set up Newton's second law for the $\hat \Theta$ direction: $$\sum F_\theta = m a_{\theta}$$
You can make use of the general expression for $a_{\theta}$ that you posted in your first post.

I am a bit lost with the $\hat \Theta$ direction, but should it be: $$-Mg = m[(-b\dot Θ^2) \hat r + b \ddot Θ \hat Θ]$$
How do I show that $\dotΘ$ is constant? If it is constant then $\ddot Θ=0$ and the equation becomes $-Mg = m(b\dot Θ^2 \hat r)$

 

[TSny]

I am a bit lost with the $\hat \Theta$ direction, but should it be: $$-Mg = m[(-b\dot Θ^2) \hat r + b \ddot Θ \hat Θ]$$

You want to apply $$\sum F_{\Theta} = m a_{\Theta}$$ The left side is the sum of all of the $\Theta$-components of the forces acting on P. Do any of the three forces that act on P have a nonzero $\Theta$ component?

On the right side of the equation, you want to write out $m a_{\Theta}$.
$a_{\Theta}$ is just the $\Theta$-component of $\vec a$. Which part of $\vec a = (-b\dot Θ^2) \hat r + b \ddot Θ \hat Θ$ represents $a_{\Theta}$?

 

[Nicci]

You want to apply $$\sum F_{\Theta} = m a_{\Theta}$$ The left side is the sum of all of the $\Theta$-components of the forces acting on P. Do any of the three forces that act on P have a nonzero $\Theta$ component?

I do not see a nonzero $\Theta$ component on the three forces. So that means that $\sum F_Θ = 0$
The Θ-component of $\vec a$ is the second part: $b\ddot Θ \hat Θ$
Then $\sum F_Θ = 0 = m\vec a_Θ = m(b\ddot Θ \hat Θ)$

 

[wrobel]

and what about period of small oscillations near the stationary rotation? :)

 

[TSny]

$0 = m\vec a_Θ = m(b\ddot Θ \hat Θ)$

Good. Can you determine anything about $\ddot \Theta$ from this?

 

[Nicci]

Good. Can you determine anything about $\ddot \Theta$ from this?

Yes, $\ddot \Theta = 0$ and if it is equal to zero, then $\dot \Theta$ is a constant. So Particle P has constant speed. Now I can say that $u = b\dot \Theta$
Can I then say that the centripetal force is: $$F_C = mu^2/b$$

 

[TSny]

Yes, $\ddot \Theta = 0$ and if it is equal to zero, then $\dot \Theta$ is a constant. So Particle P has constant speed. Now I can say that $u = b\dot \Theta$

Yes

Can I then say that the centripetal force is: $$F_C = mu^2/b$$

Yes. But to see how this falls out of the analysis, set up Newton's second law for the $\hat r$ direction: $\sum F_r = m a_r$. From your expression $\vec a = (-b\dot \Theta^2) \hat r + b \ddot \Theta \hat \Theta$, what will you use for $a_r$?

 

[Nicci]

Yes
Yes. But to see how this falls out of the analysis, set up Newton's second law for the $\hat r$ direction: $\sum F_r = m a_r$. From your expression $\vec a = (-b\dot \Theta^2) \hat r + b \ddot \Theta \hat \Theta$, what will you use for $a_r$?

For $a_r$ I will use $-b\dot \Theta^2$
$\vec T = -Mg\hat r$ so that will be my 'force'.
$$-Mg = m(-b\dot \Theta^2)$$
Since $u=b\dot \Theta$ I can say that $\dot \Theta = u/b$
Then: $$Mg = mb(u^2/b^2)$$
$$Mg = mu^2/b$$
$Mgb = mu^2$ and therefore $u^2 = Mgb/m$

I finally got it!
Thank you very very Much!

 

[TSny]

For $a_r$ I will use $-b\dot \Theta^2$
$\vec T = -Mg\hat r$ so that will be my 'force'.
$$-Mg = m(-b\dot \Theta^2)$$
Since $u=b\dot \Theta$ I can say that $\dot \Theta = u/b$
Then: $$Mg = mb(u^2/b^2)$$
$$Mg = mu^2/b$$
$Mgb = mu^2$ and therefore $u^2 = Mgb/m$

I finally got it!
Thank you very very Much!

Looks good.