Tension in a string connecting two blocks having equal and opposite velocities

[mrpurpletoes]

Homework Statement

Two blocks of masses 2kg and 3kg are connected by string of length 1m. At any instant
the velocities of block of mass 2kg and 3kg is 5 m/s in opposite direction and
perpendicular to the length of string and is also parallel to horizontal table. find tension in the string. (image attached below)

Relevant Equations

mv^2/r = centripetal force

from the question i thought that the two blocks would be carrying out uniform circular motion about a on the string such that the centripetal force (tension) acting on each block would be equal:
$$ \frac{m_{1} v_{1}^{2}}{1 - x} = \frac{m_{2} v_{2}^{2}}{x} $$
and on solving for x (taking m1 as 3 and m2 as 2) got x=2/5
putting x back into
$$ \frac{m_{1} v_{1}^{2}}{1 - x}=T $$
i got tension as 125 Newton, but the correct answer is 120 Newton according to the answer key.

 

[.Scott]

If I was editing that workbook, I would mark up that question.

Those velocities are presumably relative to the table.
And "for any instant" suggests that this system of masses and string will remain on the table with those same velocities. It is in contrast to "at one instant". But that cannot happen without some external force acting on the system. The question does not suggest how that force acts. Is one of the two masses being held to an 5m/s orbit that allows the other to orbit freely a 5m/s - effectively making the system work as one with two equal masses? If so, is it acting as two equal masses of 2kg or 3kg - or is the force split between them to act as a 2.5kg+2.5kg system?

I would check to see if there were similar problems work on in class or described in the text or the workbook.

 

[Lnewqban]

from the question i thought that the two blocks would be carrying out uniform circular motion about a on the string such that the centripetal force (tension) acting on each block would be equal:

Just imagine a huge mass and a very small mass attached at each end of the string.
Then, a horizontal rotation being induced, assuming no friction on the horizontal surface supporting the assembly.

Naturally, that mass₁-string-mass₂ assembly would try to freely rotate about the location of its center of mass.
That center would be located very close to the huge mass.

In that case, could the magnitude of the tangential velocity of each mass be the same, like in our problem?

 

[kuruman]

Let's examine whether the masses undergo uniform circular motion as assumed by the OP by looking at the tension as the centripetal force acting on each mass. Noting that the speeds of the masses are equal, $$T=\frac{m_1\cancel{v_1^2} }{x_1}=\frac{m_2\cancel{v_2^2} }{x_2}\implies m_1x_2=m_1x_1.$$
But we can also write the centripetal force in terms of the common angular speed $\omega$, $$T=m_1\cancel{\omega^2}x_1=m_2\cancel{\omega^2}x_2\implies m_1x_1=m_2x_2.$$It seems that, unless the masses are equal, the assumption that they undergo uniform circular motion with equal speeds is inappropriate. I think that this problem is overdetermined.

 

[Gavran]

What about friction?

 

[kuruman]

What about friction?

If it's not explicitly mentioned or hinted at, one must assume that that it's not there.

 

[Steve4Physics]

i got tension as 125 Newton, but the correct answer is 120 Newton according to the answer key.

The system's centre of mass (CoM) has some velocity. If you find this and transform the velocities to the CoM frame, the figures are then consistent with uniform circular motion about the centre of mass.

This gives a tension of 120 N, as required.

Note: the unit 'newton' is written with a lower case 'n' even though the symbol (N) is written upper case. Also, to be pedantic, the personal pronoun is 'I' (upper case)!

 

[kuruman]

The system's centre of mass (CoM) has some velocity. If you find this and transform the velocities to the CoM frame, the figures are then consistent with uniform circular motion about the centre of mass.

This gives a tension of 120 N, as required.

Note: the unit 'newton' is written with a lower case 'n' even though the symbol (N) is written upper case. Also, to be pedantic, the personal pronoun is 'I' (upper case)!

It is not necessary to assume that the CoM has some velocity. The two masses orbit each other while the CoM is stationary. This is shown in the figure on the right (drawn to scale). For the string to be taut in a straight line at constant tension, the period of each mass around the CoM must be the same. Since the blue circle has a larger circumference than the red circle, the blue mass must have a higher speed relative to the table than the red mass. This will be the case whether the CoM is moving or not.

The key to solving this problem is in finding the angular velocity $\boldsymbol{\omega}$ first. One needs to devise a way to use the, rather deceptively given, equal speeds but opposite velocities of 5 m/s. A better way to say it is "At any instant, the velocities of the masses are in opposite directions and perpendicular to the string and the speed of one mass relative to the other is 10 m/s."

 

[Steve4Physics]

It is not necessary to assume that the CoM has some velocity.

It's not an assumption. Using the Post #1 diagram we see that the momentum of the 3 kg mass is 15 kgm/s in the -y direction and the momentum of the 2 kg mass is 10 kgm/s in the +y direction.

It’s natural (and IMO convenient) to move into the CoM frame. Then only simple calculations are needed to:
a) check that the 2 angular speeds about the CoM are the same (i.e. do a consistency-check on the data);
b) find the tension.

Of course other approaches are possible.

Edit - typo's.

 

[kuruman]

It's not an assumption. Using the Post #1 diagram we see that the momentum of the 3 kg mass is 15 kgm/s in the -y direction and the momentum of the 2 kg mass is 10 kgm/s in the +y direction.

This means that the velocity of the CM at the moment shown in the diagram is $$\mathbf V_{\text{cm}}=\frac{-15+10}{3+2}(\text{m/s})~\mathbf{\hat y}=-1~(\text{m/s})~\mathbf{\hat y}.$$Will the CM velocoty be pointing in the same direction some time later, say a quarter period? If not, what external force accelerates the CM?

 

[Steve4Physics]

This means that the velocity of the CM at the moment shown in the diagram is $$\mathbf V_{\text{cm}}=\frac{-15+10}{3+2}(\text{m/s})~\mathbf{\hat y}=-1~(\text{m/s})~\mathbf{\hat y}.$$Will the CM velocoty be pointing in the same direction some time later, say a quarter period?

Yes. The system described consists of 2 masses connected by a string with no net external forces. So the CM has a constant velocity of $-1~(\text{m/s})~\mathbf{\hat y}$.

BTW I'm assuming the original question is badly worded so that:
"At any instant the velocities [are 5 m/s in opposite directions]”
should be:
"At some instant the velocities [are 5 m/s in opposite directions]”
The speeds could not remain equal - otherwise the system, as described, would be behaving unphysically.

We have a binary system - uniform circular motion with unequal masses orbiting their CM. In the CM frame:

If the system is also translating at 1 m/s in the -y direction, then the speeds (at the instant depicted) are both 5 m/s, corresponding to the original question.

However, I like your approach using the relative velocities/speeds of the two masses which is more general (thank you for the PM).

 

[Gavran]

If the system is also translating at 1 m/s in the -y direction, then the speeds (at the instant depicted) are both 5 m/s, corresponding to the original question.

The centre of mass circulates with a constant velocity of 1 m/s and a radius of 0,1 m.

 

[kuruman]

The centre of mass circulates with a constant velocity of 1 m/s and a radius of 0,1 m.

And what external entity is there to provide the centripetal acceleration to the center of mass? Please look at the figure in post #8.

 

[Gavran]

And what external entity is there to provide the centripetal acceleration to the center of mass? Please look at the figure in post #8.

The force of tension provides the centripetal acceleration to the centre of mass.

 

[Steve4Physics]

The force of tension provides the centripetal acceleration to the centre of mass.

The centre of mass is some point on the string. At this point, 2 forces act:
- tension: 120 N towards the 3 kg mass;
- tension: 120 N towards the 2 kg mass.
Since these forces have equal magnitudes and opposite directions, the net force = 0

Tension in the string is an internal force of the system. Internal forces do not affect the motion of a system's centre of mass.

Edit - typo'.

 

[Gavran]

There is a problem with wording. In the post #14 I mean the centripetal acceleration of masses m₁ and m₂ which is directed to the centre of mass.
There is an external entity which provides the circular motion of the centre of mass. Maybe the external entity is the circular motion of the horizontal table.

 

[jbriggs444]

There is a problem with wording. In the post #14 I mean the centripetal acceleration of masses m₁ and m₂ which is directed to the centre of mass.
There is an external entity which provides the circular motion of the centre of mass. Maybe the external entity is the circular motion of the horizontal table.

You will need to be more specific about this.

Are you suggesting that the midpoint of the string is pinned to the table? So that the tensions in the two half-strings are different.

Or are you suggesting that the table is equipped with circular frictionless rails? So that the tension in the string is irrelevant.

Or are you suggesting that the table is ruled like graph paper and moves so that the coordinate speed of both masses is always 5 m/s and the two coordinate velocities are always equal and opposite? So that for all we know the two masses are motionless with the table rotating under them.

 

[kuruman]

There is a problem with wording. In the post #14 I mean the centripetal acceleration of masses m₁ and m₂ which is directed to the centre of mass.
There is an external entity which provides the circular motion of the centre of mass. Maybe the external entity is the circular motion of the horizontal table.

The real problem with wording is in the statement of the question. The masses cannot be unequal and have equal speeds in opposite directions. An observer at rest with respect to one mass sees the other mass orbit around him with some frequency $\omega$. The same can be said about an observer at rest with respect to the other mass. From the standpoint of an inertial frame, the two masses orbit about the center of mass as shown in the figure on the right. It's like a binary system of two stars of unequal masses with tension in the string replacing the gravitational force.

The blue mass is smaller and farther away from the CM than the red mass. The condition that says this is the definition of the CM, in this case $m_1x_1=m_2 x_2.$ Now the orbital speed of each mass about the CM is given by $v=\omega x.$ If you multiply both sides of the condition equation by $\omega$, you get $$m_1 \omega x_1=m_2 \omega x_2 \implies m_1v_1=m_2 v_2.$$ Since the velocities associated with these speeds are in opposite directions, this says that the total momentum of the binary system is zero, therefore the CM does not move. It also says that the speeds cannot be equal unless the masses are equal in which case the blue and red orbits coincide.

The figure that comes with the problem is misleading but not hopelessly wrong. If one solves the problem from the standpoint of an observer in the non-inertial frame of one of the masses, the answer comes out to be 120 N as reported by the OP.

 

[Orodruin]

The real problem with wording is in the statement of the question. The masses cannot be unequal and have equal speeds in opposite directions.

Of course they can, but not at all times.

 

[kuruman]

Of course they can, but not at all times.

Sure, but the screenshot with the statement of the problem in post #1 specifies "at any instant" and we have to go with that.

 

[Orodruin]

Sure, but the screenshot with the statement of the problem in post #1 specifies "at any instant" and we have to go with that.

The problem statement also states different masses with the same velocity. So we have to go with that? But those two are incompatible. As stated, something in the problem statement needs to be corrected. The simplest change is striking out the y in ”any”, but we simply cannot know what was the intention of the author.