# Question on particle dynamics

1. Mar 22, 2012

### nik jain

1. The problem statement, all variables and given/known data
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)

3. 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) .

#### Attached Files:

• ###### IMAGE1.jpg
File size:
1.9 KB
Views:
107
2. Mar 22, 2012

### 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?

3. Mar 23, 2012

### 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 ?

4. Mar 23, 2012

### tiny-tim

(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
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.