IIT JEE , AIEEE Forum - Physics , Mechanics

atavistic

A solid cube of wood of side 2a and mass M is resting on a horizontal surface.The cube is constrained to rotate about the axis AB only.See fig.A bullet of mass m is shot with velocity v at the face opposite of ABCD at a height 4a/3.The bullet is embedded in the cube.Find the min. velocity v so as to tip the cube so that it falls on face ABCD. Assume M>>>m.

ramyani

indeed a gud Q.

hsbhatt

Part 1: Calculation of the initial angular velocity

Conservation of angular momentum about the hinge gives 4mva/3 = I

I = 2Ma²/3+2Ma² (Parallel axes theorem)

= 8Ma²/3

Hence

_o = mv/2Ma

Part 2: Condition for not tipping over

The block now begins to rotate about side AB. If the block is to not tip over, then when the centre of mass of the block is vertically above side AB, the angular velocity is zero.

Since the work done by non-conservative forces at the hinge is zero, we can apply the energy conservation law as KE_i = PE_f

1/2 I

_o² =

2 Mga

Hence 1/3 m²v²/M =

2 Mga

Hence the min velocity to ensure the block tips over is M/m

6ga

I guess this is the answer that will appear in the text book. Only I am not comfortable with this business of conserving angular momentum at the beginning. Someone pls clarify how it is possible to apply it in this case.

atavistic

Wrong answer.

ramyani

the condn of toppling is

F ( 4a/3 ) > Mga [ M >> m ]

but then how to calculate F ?

or, we hav to take a different approach ?

jasss

hsbhatt angular momentum about the hinge is not conserved in this case as we are not talking about a gravity free space .even if the collision forces cancel out N and mg do not cancel out.when the block finally falls down on ly then N&Mg cancel out but at that instant

=0.L=0

jasss

dat's correct ramayani------ F*4/3>Mg

F>3/4 Mg
now this force acts on the opposite side on d bullet.

-3/4Mg=ma....a is the deceleration of the bullet once it enters the block.
v(f)=0;
v(f)=v(i)+at
v(f)=0;

-3/4 Mgt/m=v....but problem is we dont know the time

ramyani

call big Bs like greatdreams, karthik .

jasss

Normal reaction is a non-conservative force..

shashankparewar

the velocity of the centre of mass after coll is mv/M+m
the centre of the mass is at the distance of root2*a from axis of rot.

from this you can find the angular velocity.

after that find the impulse imparted by ther particle m.

from axis
find the torque of the impulse which is the angular impulse.
equate it with the change in ang mom.you will get the answer.

konichiwa2x

Is the answer

?

atavistic

Answer is M/m[3ga(^{[ 2]}

2 -1)]^1/2

I tried CoM and energy approach.But I am not getting the answer,I am getting a 2 instead of the 3 there.

What i did was initial height of CoM(y)= a,since M>>>m so final also a,now i rotated the cube so that its base makes angle of 45 with the horizontal.This way the CoM goes to a height ^{[2 ]}

2a.So change in GPE = (M+m)ga(^{[ 2]}

2 - 1) this should be equal to KE which I used as 1/2mv^2...maybe u have to use rotational KE but i dont know inertia about that axis.Plz help someone.

The problem is I havent used 4a/3 anywhere :S

vijeshbhute

let w be the angular velocity of cube just after collision about an axis AB.As axis is taken at AB force due to hinge is cancelled and taking bullet and block as system, we can conserve angular momentum. Also the impulse due to gravity is negligible during collision.
mv*4a/3=I(about AB)*w
I(about AB)=I(com)+mr*r
(r is perpendicular dist. of axis through AB from com r=a{2}^1/2)
4mva/3=[M{4a*a+4a*a}/12+M*{2a}{2a}]*w
4mva/3=8Ma*a*w/3
w=mv/{2Ma}
cube will topple if com is at a height of a{2}^1/2
0.5I*w*w=Mga{(2)^1/2-1)
0.5*8Ma*a/3{0.5Mv/{Ma}}^2=Mga{(2)^1/2-1)
solving gives the required answer.

vijeshbhute

please rate me.

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