Particle Dynamics: Proving t(a) < t(b) for Two Masses Connected by a String

[nik jain]

Homework Statement

Two blocks A and B , each of same mass m are attached by a thin inextensible string through an ideal pulley .
Initially block B is held in position as shown in figure . Now the block B is released . Block A will slide to right and hit the pulley in time t(a) . Block B will swing and hit the surface in time t(b).
Assume the surface as frictionless.
Then prove that ,
t(a)<t(b)

The Attempt at a Solution

Let the tension in the string be T .
L=1/2(T/m)[t(a)^2]
How to get the time t(b) .

 

[tiny-tim]

hi nik jain!

the tension isn't constant, is it?

call the tension T(t),

do F = ma for block A, and F = ma and τ = Iα for block B …

what do you get? ​

 

[nik jain]

Yes Tension is not constant.

Let the block moves a distance (x) .
1/2m[v(a)^2] = Tx (By energy conservation)
1/2m[v(a)^2] = mAx (A is the acceleration of block A)
1/2m[v(a)^2] = mv(a)(dv/dx)x
integrating it
v(a) = (2x)^1/2
again integrating it
t(a) = (L/2)^1/2

How to get the t(b) ?
Please make the equation as above .
And how the acceleration of block A relates to the acceleration of block B ?

 

[tiny-tim]

1/2m[v(a)^2] = Tx (By energy conservation)

(try using the X² button just above the Reply box )

But T is not constant, so it's 1/2m[v(a)^2] = ∫ T dx

1/2m[v(a)^2] = mAx (A is the acceleration of block A)

But now you've lost T.

The whole point is to connect the A and B equations, using T.

For B, you will need two parameters, length and angle. ​