Newton's Law: Relation b/w tA & tB

[Titan97]

Homework Statement

Two Blocks A and B, each of mass M are attached by a thin inextensible string through a frictionless pulley. Block B is released from rest. Block A hits the pullty in time t_A and block B hits the vertical surface at time t_B. Find the relation between t_A and t_B.

Homework Equations

$F=ma$

The Attempt at a Solution

Length of the string from pulley to block B increases as it moves down. If the string does not become slack, and since its inextensible, I think t_A=t_B since acceleration of both ends of string should have equal acceleration. But I think its wrong. (I have attached the solution given by the author, but I want to solve it myself)
Iet block B descend by a distance $y$ and cover an angle $\theta$. Let the length of string be $R$ which can be expressed as a function of $y$ and $\theta$.
$R=ycosec\theta$.
For block A, $T=Ma$.
For block B, along y-direction, $Mg-Tsin\theta=Ma_y$ and $Tcos\theta=Ma_x$
From this, substituting value of T with Ma, $a_x=acos\theta$ and $a_y=g-asin\theta$
Then, net acceleration of Block B is $a_B=\sqrt{a^2+g^2-2agsin\theta}$
$|2agsin\theta|\le 2ag$.
But now, acceleration of both ends are different.
Here is the solution given in the textbook.

 

[Qwertywerty]

What is the relation between the acceleration of the block B along the string with that of block A's acceleration ?

 

[Titan97]

They must be equal...

 

[Qwertywerty]

And block B has acceleration along string , and perpendicular to string , so - ?

 

[Titan97]

It won't move along a circle.

 

[Qwertywerty]

And acceleration of block B with respect to block A would be - ?

 

[Titan97]

Why should i do that? I already found acceleration of block B

 

[Qwertywerty]

I have been trying to establish that acceleration of block A is lesser than that of block B , from -

What is the relation between the acceleration of the block B along the string with that of block A's acceleration ?

And block B has acceleration along string , and perpendicular to string , so - ?

 

[Chestermiller]

As your analysis stands now, your kinematic relationships between the accelerations of blocks A and block B are incorrect.

Have you been learning about using cylindrical coordinates yet? If so, this problem is much easier to solve using cylindrical (r-θ) coordinates, with the origin of the coordinate system for block B located at the pulley. Do you know the equations for the radial acceleration and the angular acceleration if both r and theta are both changing with time?

Let r be the radial distance of block B from the pulley and let theta be the same theta you have been using.

Chet

 

[PeroK]

Homework Statement

Two Blocks A and B, each of mass M are attached by a thin inextensible string through a frictionless pulley. Block B is released from rest. Block A hits the pullty in time t_A and block B hits the vertical surface at time t_B. Find the relation between t_A and t_B.
View attachment 87385

Homework Equations

$F=ma$

The Attempt at a Solution

Length of the string from pulley to block B increases as it moves down. If the string does not become slack, and since its inextensible, I think t_A=t_B since acceleration of both ends of string should have equal acceleration. But I think its wrong. (I have attached the solution given by the author, but I want to solve it myself)
Iet block B descend by a distance $y$ and cover an angle $\theta$. Let the length of string be $R$ which can be expressed as a function of $y$ and $\theta$.
$R=ycosec\theta$.
For block A, $T=Ma$.
For block B, along y-direction, $Mg-Tsin\theta=Ma_y$ and $Tcos\theta=Ma_x$
From this, substituting value of T with Ma, $a_x=acos\theta$ and $a_y=g-asin\theta$
Then, net acceleration of Block B is $a_B=\sqrt{a^2+g^2-2agsin\theta}$
$|2agsin\theta|\le 2ag$.
But now, acceleration of both ends are different.
Here is the solution given in the textbook.
View attachment 87390

That solution makes no sense. Instead, you could go back to your initial approach. Hint: You can solve this without any calculation.

Perhaps think about the cases where block B is much heavier or much lighter than block A. What happens in those cases?

Then you might see what happens when the two blocks are of equal mass.

 

[Titan97]

Firstly @Chestermiller , I only know the definition of cylindrical coordinates. But I will try to use it.
Secondly, @PeroK , If block B is much heavier than block A, the time it takes to fall down won't change right? A piece of feather and a block of iron take the same time to fall in vacuum. If block A is heavier, since there is no friction, I still don't think there will be any change.

 

[PeroK]

Firstly @Chestermiller , I only know the definition of cylindrical coordinates. But I will try to use it.
Secondly, @PeroK , If block B is much heavier than block A, the time it takes to fall down won't change right? A piece of feather and a block of iron take the same time to fall in vacuum. If block A is heavier, since there is no friction, I still don't think there will be any change.

If block B is much heavier then it will fall straight down and pull block A into the pulley.

If block B is much lighter, it will swing into the wall like a pendulum and A will hardly move.

The question is, if they are the same mass, which one gets pulled more, or do they get pulled equally?

 

[Titan97]

@PeroK , here is how I tried to solve it. The force acting along horizontal for block B is $Tsin\theta$ and the force acting along horizontal on block A is $T$.
$Tsin\theta\le T$. Hence, the acceleration along horizontal direction for block A is more and hence it hits the pulley first or takes lesser time.

 

[Chestermiller]

@PeroK , here is how I tried to solve it. The force acting along horizontal for block B is $Tsin\theta$ and the force acting along horizontal on block A is $T$.
$Tsin\theta\le T$. Hence, the acceleration along horizontal direction for block A is more and hence it hits the pulley first or takes lesser time.

I'm embarrassed to admit that I tried a much more complicated approach to this problem and was not able to reach a conclusion. That's because I violated my own first three most important rules of modeling: 1. keep it simple, 2. keep it simple, 3. keep it simple.

Nice job Titan97.

Chet