Laws of Motion - Basic Problem

1. Jun 17, 2012

draotic

1. The problem statement, all variables and given/known data
Given in the pic .. .

2. Relevant equations
F=mg

3. The attempt at a solution
i think block 2 hitting wedge will take place 1st as the direction of displacement of B1 is perpendicular to 'g' .. .
but i dont think thats a logical answer

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2. Jun 17, 2012

tiny-tim

hi draotic!

call the displacement "x" and the angle "θ" …

what equations do you get?

3. Jun 18, 2012

draotic

sorry , i didnt quite get it .. .
which angle are we concerned about?

4. Jun 18, 2012

tiny-tim

the angle the right-hand part of the string makes with the horizontal at any particular time

5. Jun 18, 2012

draotic

well the angle will vary from 0 to pi/2 , so are we going to involve integration ?

6. Jun 18, 2012

azizlwl

Block#2 needs the highest force up at vertically down, weight plus centripetal force before continuing to hit the wedge.

Thus block#1 must hit the pulley first before bigger block can continue the circular motion.

7. Jun 19, 2012

tiny-tim

hi draotic!

(just got up :zzz:)
in problems like this, always try an energy equation first (which will avoid integration)

if that doesn't work, use a force equation (which will)

8. Jun 19, 2012

draotic

thanks tiny-tim and azizlwl for helping me out ..
azizlwl's reason seems quite right to me..
in B1 , only tension is involved ( ignoring kinetic friction ) and in B2 , we must have tension + centripetal force . .

so B1 must hit pulley first..

9. Jun 19, 2012

draotic

thanks tiny-tim and azizlwl for helping me out ..
azizlwl's reason seems quite right to me..
in B1 , only tension is involved ( ignoring kinetic friction ) and in B2 , we must have tension + centripetal force . .

so B1 must hit pulley first..

10. Jun 19, 2012

nrbhyagrwl

are all the parameters such as masses, distances from pulley variable?

11. Jun 19, 2012

draotic

masses are same , no friction anywhere , no MI of pulley

12. Jun 19, 2012

tiny-tim

you can assume the pulley is massless

(alternatively, just add its "rolling mass" (I/r2) to the mass of the first block)