Question on particle dynamics

  • Thread starter nik jain
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  • #1
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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) .
 

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Answers and Replies

  • #2
tiny-tim
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hi nik jain! :smile:

the tension isn't constant, is it? :wink:

call the tension T(t),

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

what do you get? :smile:
 
  • #3
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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
tiny-tim
Science Advisor
Homework Helper
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1/2m[v(a)^2] = Tx (By energy conservation)

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

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. :redface:

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

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

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