23
qwerty
·2010-11-25 22:22:40
a1 is accn of m in grnd frame ,
accn of m wrt M will be a1 - a2
1
swordfish
·2010-11-26 03:14:51
Hey how is a1 w.r.t ground? I am working from the frame of reference attached on the block M, so there is a pseudo force acting on m towards right, friction towards left. Whatever accln. I find from the frame attached to M will be w.r.t to M right?
71
Vivek @ Born this Way
·2010-11-26 08:47:21
I have really doubt in this question.
The system is started towards right with an initial velocity v
How is Pseudo force acting in right then?
23
qwerty
·2010-11-26 09:20:44
swordfish u hav written
"ma1 = μmg/2"
this is eqn for m in grnd frame or in M's frame ?
1
swordfish
·2010-11-26 10:56:06
Errrm...may be in ground frame......
What will be its equation in the frame of M then?
One doubt...how do you find whether the given equation is from ground frame or frame of M.
@ Vivek...
Think of it this way.......The system is initially started with a velocity v ( assume no friction present....for time being)...........and then the whole system is kept on the surface where friction is present.....so there is a retardation acting on the block M.
6
Kalyan IIT-K Beware I'm coming
·2010-11-26 19:41:49
isnt this the hcv sum?
anyways
no sawal of pseudo force here
it says the system
just remember
friction is self-adjustable
:P:P
1
swordfish
·2010-11-27 01:04:48
No sawal of pseudo force?
Then what causes the smaller block to move forward and finally fall? Some kind of Devilish Force? haha
23
qwerty
·2010-11-27 03:12:25
there is nothing like "pseudo force" in reality
btw swordfish it is better if u accelerations of each m in ground frame and then find relative accn of m wrt M, and then use that kinematics eqn ...
1
swordfish
·2010-11-27 05:10:43
Look for the block M, its acceleration w.r.t ground will be along -i (it is decelerating)
\mu (m+M)g(-\hat{i})+\frac{\mu}{2}mg(\hat{i})=M\vec{a}
\vec{a}=\mu g\left(1+\frac{m}{2M}\right)(-\hat{i})
Now when we observe the block m from ground, it accelerates forward with
\vec{a_1}=\left(\frac{\mu}{2}\right)g(\hat{i})
Now relative acceleration of m w.r.t ground gives wrong answer.
Please help qwerty.
23
qwerty
·2010-11-28 05:25:05

so m will accelerate along - ve x , ( but its accn will be less than accn of M bcz of given value of μ, hence it is able to cover that distance L )
MaM =μmg + μMg - μmg/2
so aM = μg( 1 + m2M) (-i)
am = μ2g ( -i )
am/M = μg( 1 + m2M) - μ2g ( i ) = μg2(1+mM) (i)
so
sm/M = um/M+12am/Mt2
L = 0 + μg4(1+mM)t2
t2 = 4LMμg(m+M)
1
swordfish
·2010-11-28 08:47:11
I just wanted this thing-
so m will accelerate along - ve x , ( but its accn will be less than accn of M bcz of given value of μ, hence it is able to cover that distance L )
There was no need of solution. Im so sorry for wasting your time.
Thanks alot!!!