comprehension

anyone help na ???

Continuation of my previous post....

The FBD of the bob m:

Since m is moving in a circle, the centripetal acceleration is supplied by components of N and mg. So we get
N\cos\theta+mg\sin\theta = m\omega^2\ell
When it leaves contact with the block, N = 0 and θ = π/6. First, at θ = π/6, we get
\omega^2 = \dfrac{g}{\ell}\cdot \dfrac{1}{1+\frac{M}{4m}}
Hence, we have by setting N = 0, we get
\dfrac{g}{2} = g \dfrac{1}{1+\frac{M}{4m}}
i.e.
\boxed{\dfrac{M}{m}=4}

When the rod makes an angle θ with the horizontal, let the velocity of the block M be V while its acceleration be a_M. At this instant, let the angular velocity of the rod about the hinge A be ω.

Since the bob is in contact with the block, the horizontal velocity of the bob must be V. Therefore
\omega \ell\sin\theta =V
Next, energy conservation gives
mg\ell(1-\sin\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}m\ell^2\omega^2
Eliminating V gives
mg\ell(1-\sin\theta)=\dfrac{1}{2}M\omega^2\ell^2\sin^2\theta+\dfrac{1}{2}m\ell^2\omega^2=\dfrac{1}{2}\omega^2\ell^2(m+M\sin^2\theta)
which gives
\omega^2=\dfrac{2mg(1-\sin\theta)}{\ell(m+M\sin^2\theta)}
Differentiate w.r.t. t and use the fact that \omega=-\dfrac{\mathrm d\theta}{\mathrm dt}, we get the angular acceleration \dfrac{\mathrm d\omega}{\mathrm dt}=\dfrac{g}{\ell}\cdot\dfrac{m\cos\theta(m+M\sin\theta(2-\sin\theta))}{(m+M\sin^2\theta)^2}

Hence the tangential acceleration of the bob is
a_t=g\cdot\dfrac{m\cos\theta(m+M\sin\theta(2-\sin\theta))}{(m+M\sin^2\theta)^2}
Also, the acceleration a_M of the block is
a_M=\dfrac{\mathrm dV}{\mathrm dt}=\ell\left(\sin\theta \dfrac{\mathrm d\omega}{\mathrm dt}+\omega \cos\theta\dfrac{\mathrm d\theta}{\mathrm dt}\right)=\ell\left(\sin\theta \dfrac{\mathrm d\omega}{\mathrm dt}-\omega^2 \cos\theta\right)
Solving for the tangential acceleration a_t in terms of a_M and V we get
\boxed{a_t = \dfrac{a_M\sin\theta + (V^2/\ell)\cos\theta}{\sin^2\theta}}
as the answer to the first part.

shortcut for relation between vel

vel of m is wl
w=d@/dt i think this should be enugh!

Q1 is my dbt....plz help

correct ans kya hai ???
sriraghav can u plzz post the ans also ?? not the soln, but the just the final ans ...

ok then post ur method and ans i see now it doesnt depend on v

N wouldnt be a constant

it would depend on the vel i think but yet not solved fully

but what about my doubt in the 3rd post

right
cmon how wud you solve anything if it goes to left

i have another doubt with the question

(and this answers ur doubt i think philip)

the small mass goes to the left or right when it is jerked????

wheres this point A

can someone explain why or how velocity of mass M is needed for first part... [7][7]

on ly forces that'll contribute to tangential acceleration will be mg and N na???

someone plz see this