Why doesn't the released mass m2 on pulley have higher acceleration?

[annamal]

TL;DR

Why doesn't the released mass m2 on pulley have higher acceleration since it is no longer subject to the upward force of tension?

Consider the pulley in the attached image to be frictionless. (a) If m2 is released, what will its acceleration be?

My question is why wouldn't m2's acceleration be greater if released rather than attached to the string because m2 released is no longer subject to the string's upward force tension?

 

[berkeman]

Summary:: Why doesn't the released mass m2 on pulley have higher acceleration since it is no longer subject to the upward force of tension?

Consider the pulley in the attached image to be frictionless. (a) If m2 is released, what will its acceleration be?

My question is why wouldn't m2's acceleration be greater if released rather than attached to the string because m2 released is no longer subject to the string's upward force tension?

Is there an actual full problem statement associated with this image? Of course if you cut the string, M2 will free fall. But are they actually asking about when the 2 masses are still connected to the string? Then the acceleration will depend on the difference between the 2 masses, no?

 

[annamal]

Is there an actual full problem statement associated with this image? Of course if you cut the string, M2 will free fall. But are they actually asking about when the 2 masses are still connected to the string? Then the acceleration will depend on the difference between the 2 masses, no?

That's how the problem is worded. I think it just means at that instant the mass 2 is released.

 

[berkeman]

That's how the problem is worded. I think it just means at that instant the mass 2 is released.

Usually "released" means let go out of your hand. If they were cutting the string, the verb "cut" would likely be used. I suppose you could submit both answers for extra credit to your teacher...

 

[Lnewqban]

"Consider the pulley in the attached image to be frictionless. (a) If m2 is released, what will its acceleration be?"

If the problem would mean to cut the string, how important the friction in the pulley would be?
As @berkeman stated above, somebody has been holding mass 2, preventing the system from moving under its own forces.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

 

[annamal]

"Consider the pulley in the attached image to be frictionless. (a) If m2 is released, what will its acceleration be?"

If the problem would mean to cut the string, how important the friction in the pulley would be?
As @berkeman stated above, somebody has been holding mass 2, preventing the system from moving under its own forces.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

a) If m2 were cut, initially a2 = a (acceleration of mass on string) - g but then a2 grows to equal -g
Is that correct?

 

[jbriggs444]

a) If m2 were cut, initially a2 = a (acceleration of mass on string) - g but then a2 grows to equal -g
Is that correct?

As I understand this new question, you previously have released mass $m_2$ from your hand. It is accelerating at some rate $a$ while mass $m_1$ is accelerating at the same rate in the opposite direction.

You ask about how the acceleration changes from $a$ to $-g$.

If you snip the [ideal, inextensible] cord above mass $m_2$ instantly then the acceleration of $m_2$ will change instantly. It will not slowly "grow".

 

[vela]

a) If m2 were cut, initially a2 = a (acceleration of mass on string) - g but then a2 grows to equal -g
Is that correct?

No. Use $F=ma$. What forces are on $m_2$ after the string is cut? That's what determines the acceleration.

 

[annamal]

No. Use $F=ma$. What forces are on $m_2$ after the string is cut? That's what determines the acceleration.

but isn't the acceleration of the cut mass going to be influenced by the acceleration the mass is going with the string

 

[annamal]

As I understand this new question, you previously have released mass $m_2$ from your hand. It is accelerating at some rate $a$ while mass $m_1$ is accelerating at the same rate in the opposite direction.

You ask about how the acceleration changes from $a$ to $-g$.

If you snip the [ideal, inextensible] cord above mass $m_2$ instantly then the acceleration of $m_2$ will change instantly. It will not slowly "grow".

can this instant change be measured in time of milliseconds or even less?

 

[annamal]

can this instant change be measured in time of milliseconds or even less?

that confuses me because doesn't it take time to change acceleration?

 

[Lnewqban]

a) If m2 were cut, initially a2 = a (acceleration of mass on string) - g but then a2 grows to equal -g
Is that correct?

Sorry, I don't understand.
My previous post only tried to reason about the release of mass 2 (from a repose condition), rather than being allowed to free fall (detached from the string and the effect of mass 1).

If mass 2 is released, tension in the string increases to a maximum value, which reaches mass 1.
After that instant, mass 1-string-mass 2 become a system that moves (or not) at same rate of acceleration.
Such a system is also known as an Atwood machine.

Being frictionless, the pulley only job here is to change the direction of the string and its internal tension.

 

[jbriggs444]

that confuses me because doesn't it take time to change acceleration?

If the force changes discontinuously then so does the acceleration.

Edit: Yes, there are always (almost always?) limits on how rapidly forces can be changed. But they can be changed pretty rapidly.

 

[annamal]

If the force changes discontinuously then so does the acceleration.

Edit: Yes, there are always (almost always?) limits on how rapidly forces can be changed. But they can be changed pretty rapidly.

So therefore if the mass is accelerating at a1 attached to the string, if the string were cut, it would first accelerate at a1 - g and then -g. correct?

 

[jbriggs444]

So therefore if the mass is accelerating at a1 attached to the string, if the string were cut, it would first accelerate at a1 - g and then -g. correct?

Sorry, egg all over my face. Forgot which thread I am on.