Final speed when the mass hits the post

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

A) Assuming no gravity
The only external force acting on the mass m is tension and the torque due to it about the axis of post is 0.
Hence, applying conservation of angular momentum about the axis of post gives,
## m v_0 r = m v R
\\ v = v_0 r/R ##
B)The only external force acting on the mass m is tension and the torque due to it about the axis of post is RT ##\hat z##.
I can't say the conserved quantity in advance.
Taking ##\theta ## to be the polar angle,
## r(\theta ) = r_i - R \theta ##
Work energy theorem tells us that net work done on it = change in kinetic energy
## W = \int _{\theta _i} ^{ \theta _f} RTd\, \theta##
## \vec a = (\ddot r - r {\dot \theta}^2 ) \hat r + (2 \dot r \dot \theta + r \ddot \theta ) \hat \theta ##
## T = m (\ddot r - r {\dot \theta}^2 )##
Is this correct so far?

 

[scottdave]

In case (b), what is preventing the post from spinning? The tension is always tangent to the radius of the post.

 

[Hiero]

I will assume the post is fixed in the ground, so cannot rotate. (Equivalently, let the post be free but take the limit of large post-mass.) Otherwise we need to know the mass of the post, and also there would be a question of if the mass ever hits the post (for small post-mass I would think not). @scottdave

Taking ##\theta ## to be the polar angle,
## r(\theta ) = r_i - R \theta ##

Taking the polar angle of? The location of the mass? Be careful here. Rθ is* the amount of string which has wrapped around the post... Is this the same as the distance the mass has moved towards the center? (This r(θ) equation says, yes.)

*[Edit: This is actually not even the correct amount of string wrapped around... for that, you would use the angle of the contact point, which is not generally the angle of the mass (only when r=R). Your method requires extra care!]

## \vec a = (\ddot r - r {\dot \theta}^2 ) \hat r + (2 \dot r \dot \theta + r \ddot \theta ) \hat \theta ##
## T = m (\ddot r - r {\dot \theta}^2 )##

Is the tension pointing purely radially? (This T equation says, yes.)Anyway, I have pointed out some inconsistencies in your method, but I do not really want to suggest that you fix them. Pursuing that approach seems very messy... There must be a better way...

Strings are complicated things in real life. Create this problem with a rope and with a bungee cord and you will observe quite different motions. In the realm of practice-problems, we like to idealize things. How do we idealize a string? What is different between the rope and the bungee cord? The stiffness, of course. A rope is idealized by saying that is inextensible. Like frictionless surfaces, massless pulleys, rigid bodies, etc... inextensible strings do not really exist in the real world. But like those other things, they are very useful approximations in some situations.

So if we take these strings to be inextensible, can the mass move along the direction of the rope (and therefore the direction of tension) in part (b)? What about in part (a)? What are the implications?

 

[haruspex]

A) Assuming no gravity

Since gravity acts vertically and you are using angular momentum about a vertical axis, you do not need that assumption here. You do need it in B).

Work energy theorem tells us that net work done on it = change in kinetic energy

Good start... is anything doing work on it?

 

[Pushoam]

Since gravity acts vertically and you are using angular momentum about a vertical axis, you do not need that assumption here. You do need it in B).

But there is a perpendicular distance between where the gravitational force is acting i.e. position vector of m and the axis of rotation.
In the presence of gravity, there will be a net torque about the axis of post acting on the mass ## \tau = r \hat s \times mg(-\hat z) = mgr \hat \phi ##
So, the conservation of angular momentum could not be applied.

 

[haruspex]

But there is a perpendicular distance between where the gravitational force is acting i.e. position vector of m and the axis of rotation.
In the presence of gravity, there will be a net torque about the axis of post acting on the mass ## \tau = r \hat s \times mg(-\hat z) = mgr \hat \phi ##
So, the conservation of angular momentum could not be applied.

Angular momentum is a vector. If there is no torque about the vertical axis, how can the component of angular momentum about the vertical axis change?
I am not claming the entire vector is conserved.

 

[Pushoam]

Can anyone please tell me how to number an eqn. in PF?Another attempt:

Using cylindrical coordinate - system, taking axis of post as z- axis and origin at the bottom of the post, and denoting by the position vector of mass m by ##\vec x,##

## \vec x = R \hat s + r \hat \phi + \vec z ##

## \dot{ \vec x} ## ##= ( R \dot \phi + \dot r ) \hat \phi ## ## - r \dot \phi \hat s + \dot {\vec z} ##

So if we take these strings to be inextensible, can the mass move along the direction of the rope (and therefore the direction of tension) in part (b)?

So, taking the above advice, component of velocity in ## \hat \phi ## is 0.
This gives ## \dot \phi = \frac {-\dot r } {R}##

## \vec F_{net} = mg (- \hat z) + T(\hat \phi)##

Can't I simplify the problem by taking no motion in ##\hat z ## direction?

You do need it in B).

Assuming no gravity,
Torque acting on the system, ## \vec \tau = RT \hat z ##
Angular displacement of the mass m is denoted by: ## d \vec \phi _s##
Now, I have to express ## d \vec \phi_s ## in terms of ## \phi ## and other known quantities and then to use work energy theorem.

Let's take ## \theta ' ## as the angle between ## \hat s ## and ## \hat s_m ## (it corresponds to the polar unit vector of mass m ).
Then ## \phi_s = \phi + \theta ' ##

No need to do any further calculation.
If I consider no motion in the ## \hat z## direction, then there is no component of ##d \vec \phi_s ## in ##\hat z ## direction as ## \hat \phi _s ## could be expressed in ## \hat \phi ## and ## \hat s## unit vectors and so is ## d \vec \phi_s ##.
Now, since ##d \vec \phi_s ## is perpendicular to ## \hat z ## , work done by torque is 0.
Since, there is no work being done on the mass, the work energy theorem says that its kinetic energy remains constant.
So, the speed ##v(t) = v_0## remains unchanged.

 

[Pushoam]

Angular momentum is a vector. If there is no torque about the vertical axis, how can the component of angular momentum about the vertical axis change?
I am not claming the entire vector is conserved.

I missed this point. Thanks for pointing it out.
In some cases the angular momentum may not be conserved, but its component in certain directions may be conserved. One can use this for simplifying the calculation.

 

[haruspex]

You did not answer my question regarding B) in post #4. If you answer it correctly you will find this case is quite easy.

 

[Pushoam]

Another question:
Is the work done by a torque equal to the work done by the corresponding force if the the corresponding coordinates remains same in both methods and the torque is calculated about the origin?
work done by a torque: ## W _{\tau } = \int_{\vec {\theta _i }} ^ { \vec{ \theta _ f }} \vec \tau . d\, \vec \theta = \int_{\vec \theta _i } ^ { \vec \theta _ f } \vec F . (\vec r \times d \vec \theta) ##
Now, in plane motion, in cylindrical coordinate system, taking ## \theta ## as polar angle,
## d \vec r = dr \hat r + r d \theta \hat \theta ##
So, in pure rotational motion with constant axis of rotation,
##(\vec r \times d \vec \theta) =d \vec r##
## W _{\tau } = \int_{\vec r _i } ^ { \vec r _ f } \vec F. d \vec r = W_{force}##

So, for pure rotational motion with constant axis of rotation,
work done by torque = work done by corresponding force, if the the corresponding coordinates remains same in both methods and the torque is calculated about the origin.

But, what is the answer for general case?But, if in general I have to calculate total work done on a rigid body system,
total work done on a rigid body system = sum of the work done by each torque + sum of the work done by each force on the center of mass

sum of the work done by each torque = change in the (rotational kinetic energy = ## \frac 1 2 I \omega ^2##)
sum of the work done by each force on the center of mass = work done by the sum of the forces i.e. net force = change in the kinetic energy of the center of mass
provided that the coordinate system and the origin should remain same for all calculation and the torque should be calculated about the origin

Is this correct so far?

 

[haruspex]

Another question:
Is the work done by a torque equal to the work done by the corresponding force if the the corresponding coordinates remains same in both methods and the torque is calculated about the origin?
work done by a torque: ## W _{\tau } = \int_{\vec {\theta _i }} ^ { \vec{ \theta _ f }} \vec \tau . d\, \vec \theta = \int_{\vec \theta _i } ^ { \vec \theta _ f } \vec F . (\vec r \times d \vec \theta) ##
Now, in plane motion, in cylindrical coordinate system, taking ## \theta ## as polar angle,
## d \vec r = dr \hat r + r d \theta \hat \theta ##
So, in pure rotational motion with constant axis of rotation,
##(\vec r \times d \vec \theta) =d \vec r##
## W _{\tau } = \int_{\vec r _i } ^ { \vec r _ f } \vec F. d \vec r = W_{force}##

So, for pure rotational motion with constant axis of rotation,
work done by torque = work done by corresponding force, if the the corresponding coordinates remains same in both methods and the torque is calculated about the origin.

But, what is the answer for general case?But, if in general I have to calculate total work done on a rigid body system,
total work done on a rigid body system = sum of the work done by each torque + sum of the work done by each force on the center of mass

sum of the work done by each torque = change in the (rotational kinetic energy = ## \frac 1 2 I \omega ^2##)
sum of the work done by each force on the center of mass = work done by the sum of the forces i.e. net force = change in the kinetic energy of the center of mass
provided that the coordinate system and the origin should remain same for all calculation and the torque should be calculated about the origin

Is this correct so far?

I'm guessing this is your answer to my question. One sentence would have sufficed, or even one word.
I'll come at it a diffent way: If the string is doing work on the mass, where is it getting the energy from?

 

[Pushoam]

You did not answer my question regarding B) in post #4. If you answer it correctly you will find this case is quite easy.

Yes, now I understood the question and the answer both.

Good start... is anything doing work on it?

Consider the string to be in-extensible, the body is not moving the direction of tension force. Assuming there is no motion in the vertical direction, work done by gravitational force is also 0.
In this case, net work done on the mass = sum of the work done by each force = sum of the work done by torque due to each force= 0
Now, by work energy theorem, kinetic energy remains unchanged and so is the speed.

 

[Pushoam]

I'm guessing this is your answer to my question. One sentence would have sufficed, or even one word.
I'll come at it a diffent way: If the string is doing work on the mass, where is it getting the energy from?

This is not answer to your question. The answer to your question is in post #12.
In post # 10, I am asking questions. And I need your help here. So, please read it.

 

[haruspex]

Is the work done by a torque equal to the work done by the corresponding force

So long as you are using the same reference frame, the work must be the same, yes.

 

[Pushoam]

So long as you are using the same reference frame, the work must be the same, yes.

But I am getting difficulty in proving it in post #10.

But, what is the answer for general case?But, if in general I have to calculate total work done on a rigid body system,
total work done on a rigid body system = sum of the work done by each torque + sum of the work done by each force on the center of mass

sum of the work done by each torque = change in the (rotational kinetic energy = 12Iω212Iω2 \frac 1 2 I \omega ^2)
sum of the work done by each force on the center of mass = work done by the sum of the forces i.e. net force = change in the kinetic energy of the center of mass
provided that the coordinate system and the origin should remain same for all calculation and the torque should be calculated about the origin

Is this correct so far?

What about this? Is this wrong?

 

[Pushoam]

The answer to your question is in post #12.

What about this?

 

[haruspex]

Now, by work energy theorem, kinetic energy remains unchanged and so is the speed.

Right.

 

[Pushoam]

Please, reply to post # 15, too.

 

[haruspex]

But I am getting difficulty in proving it in post #10.

What about this? Is this wrong?

In post #10 you correctly wrote ##\int \vec\tau.\vec {d\theta}=\int \vec F.(\vec r\times\vec {d\theta})##.
Now ##\vec r\times\vec{d\theta}=\vec {ds}##, where ##\vec {ds}## is the displacement.

 

[Pushoam]

In post #10 you correctly wrote ##\int \vec\tau.\vec {d\theta}=\int \vec F.(\vec r\times\vec {d\theta})##.
Now ##\vec r\times\vec{d\theta}=\vec {ds},## where ##\vec {ds} ## is the displacement.

In post # 10,

Now, in plane motion, in cylindrical coordinate system, taking θ as polar angle,
## \vec r = dr \hat r + r d \theta \hat \theta ##

But, ## \vec r \times d\vec \theta = - r d\theta \hat \theta ≠ d \vec r ##
So, what is wrong here?

 

[haruspex]

In post # 10,

But, ## \vec r \times d\vec \theta = - r d\theta \hat \theta ≠ d \vec r ##
So, what is wrong here?

Your original question concerned the work done by "a torque". That is not the same as the work done by a force in general.
If we take an arbitrary force and resolve it into components radial to and tangential to some chosen axis, the tangential component provides the torque. The work this component does is the same whether you calculate it as ##\vec\tau.\vec{d\theta}## or ##\vec F.(\vec r\times\vec{d\theta})##. But the radial component, which does not provide a torque, also does work if ##|\vec r|## changes.

 

[Pushoam]

Thanks for guiding me, thanks a lot