Find the initial acceleration of the block

[TSny]

Yes, That looks good to me.

 

[Vibhor]

This is a rare problem I have done where acceleration of connected particles via string are different .Generally two particles connected by a string going over a pulley have same acceleration .

 

[Vibhor]

Why doesn't $r\dot{\theta}^2$ contribute to the acceleration of B ? How should they have moved so that the accelerations would have been same ( apart from A just moving down and B moving towards right ) ?

 

[TSny]

Why doesn't $r\dot{\theta}^2$ contribute to the acceleration of B ?

The block moves horizontally along a straight line. If $x$ denotes its position along the line, then it is clear that the acceleration of the block is $\ddot{x}$. But it should also be clear that $\ddot{x} = \ddot{r}$. So, $r\dot{\theta}^2$ does not contribute to the acceleration of the block. There just isn't any reason why the accelerations of the two objects should be equal just because they are connected by a string. As an extreme example, imagine you tie a ball to a string and whirl it around in a circle such that your hand that holds the string is essentially at rest. The ball has centripetal acceleration while your hand has no acceleration even though your hand and the ball are connected by a string.

How should they have moved so that the accelerations would have been same ( apart from A just moving down and B moving towards right ) ?

I can't think of a way that they would have the same acceleration other than having A move straight downward.

 

[Vibhor]

There just isn't any reason why the accelerations of the two objects should be equal just because they are connected by a string.

Thanks for pointing this out .This was a misconception I was harboring .

So,can we say that this was a purely kinematics problem , not a dynamics one (where forces come in picture ). Of course gravity is involved here .

If B wasn't mass less ,and had the same mass (or different from A ) ,then Newton's law would come in picture . The accelerations would be determined by the forces acting on A and B .

Is this alright ??One more thing .So, it is wrong to say that A undergoes circular motion about the pulley ??

 

[TSny]

So,can we say that this was a purely kinematics problem , not a dynamics one (where forces come in picture ). Of course gravity is involved here .

Pretty much. But I would say that Newtons laws play a role in showing that the tension in the string is zero if the mass of the block is negligible.

If B wasn't mass less ,and had the same mass (or different from A ) ,then Newton's law would come in picture . The accelerations would be determined by the forces acting on A and B .

Is this alright ??

Yes.

One more thing .So, it is wrong to say that A undergoes circular motion about the pulley ??

Right, the particle does not move in circular motion about the pulley.

 

[Chestermiller]

If B wasn't mass less ,and had the same mass (or different from A ) ,then Newton's law would come in picture . The accelerations would be determined by the forces acting on A and B .

Why don't you re-do the problem for that case (using cylindrical coordinates) and see what you get? The accelerations of both blocks will still be different.

Chet

 

[Vibhor]

Why don't you re-do the problem for that case (using cylindrical coordinates) and see what you get? The accelerations of both blocks will still be different.

Chet

Sir,

I don't know how to work with cylindrical coordinates . I first need to study polar coordinates .

I have a doubt . I approximated the expression in post#22 and got the (correct) result . I was wondering how an approximated value turned out to be equal to the actual result ??

Has it anything to do with calculus ??

 

[Chestermiller]

Sir,

I don't know how to work with cylindrical coordinates . I first need to study polar coordinates .

Sorry. I tend to use the terms polar coordinates and cylindrical coordinates interchangeably.

I have a doubt . I approximated the expression in post#22 and got the (correct) result . I was wondering how an approximated value turned out to be equal to the actual result ??

Has it anything to do with calculus ??

In the limit of t = 0, the approximate result approaches the exact result (in this problem). If you wanted to, you could have differentiated d(t) twice with respect to t, and set t = 0 in the final equation and gotten the exact answer that way also.

Chet

 

[insightful]

So, rθ˙2r\dot{\theta}^2 does not contribute to the acceleration of the block.

So why does it appear in the equation for the acceleration of the block?

 

[TSny]

So why does it appear in the equation for the acceleration of the block?

You are right. Since the value of $\ddot{r}$ does depend on the value of $r\dot{\theta}^2$, the acceleration of the block does depend on $r\dot{\theta}^2$.

But the acceleration of the block is always equal to just $\ddot{r}$ of the particle rather than $a_r$ of the particle. I thought Vibhor was asking why the acceleration of the block is not equal to the complete acceleration $a_r$ of the particle (for which both $\ddot{r}$ and $r\dot{\theta}^2$ contribute).