IIT JEE , AIEEE Forum - Physics , Mechanics

shubham_sachdeva

when can we conserve angular momentum.....?? in all books its written that when torque acting on the system is 0.........

then in qns. which says that there is some sphere is resting & a bullet hits it.then find blah..blah...there tpo wewe conserve angular momentum about centre of sphere.but when bullet will strike then it will exert some torque.then why we conserved momentum.??

also see the below qn.

Q. a solid sphere is set rolling wid linear speed v . it hits the wall elastically. .now find the linear speed when it will start pure rolling ..

plzzzzz giv lots of reason's theory..

netkid07

about angular momentum i can only say that it is conserved only when NO EXTERNAL TORQUE is acting..................

in that q we take bullet n sphere as a single system so bullet only produces internal torque bcoz it is a part of system............

rate if that helped ... ... ...

shubham_sachdeva

plzzz answer the whole qn....

shubham_sachdeva

plzz sum1 else relpy......

shubham_sachdeva

common.....SOLVE................

karthik2007

See, conservation of angular momentum can be applied in case there is no external torque. Let me make it more clear by solving the sum you've given:

Consider the instant at which the ball is just about to hit the wall. It has clockwise angular momentum. As soon as it hits, and rebounds, its linear velocity becomes -v, but angular momentum is in the same direction, ie, clockwise.

So, net momentum = mvr-Iw (Assuming anticlockwise dirn to be positive)

when it attains pure rolling, net angular momentum = mv'r + Iw'.

Equating the two:

mvr - 2/5mrv = mv'r + 2/5mrv

3/5mrv = 7/5mrv'

which gives v' = 3v/7.

Here we have taken the system to be the wall and the sphere.We can apply conservation of ang momentum as there is no external torque. Don't think friction is an external torque, it is included in the system, it is because of friction that the balls roll. However, if someone kicks the ball in between, then we cant use conservation of angular momentum

shubham_sachdeva

but why friction is not an external force...similarily, gravity is an external force.....but then too we are conserving......also, mvr term will be 0.as r here is 0.bcz we are conserving about O.

priyesh

Hi shubham we conserve momentum by taking bullet & sphere as the system so that the forces become internal and we can conserve angualr & linear Momentum

REFER H.C VERMA PART 1 PG 191 Q 29 (IT'S A SOLVED EXAMPLE)

THIS WILL CLEAR UR DOUBT.

balu_don

see i will calrify before the bullet hits the sphere t=0 .we are not conserving at the time of impact after the collision thebullet sticks and there is no ext. torque that is considered as ineternal.pls rate me

priyesh

hey @karthik

FRICTION IS AN EXTERNAL FORCE SO TORQUE BY IT WOULD BE EXTERNAL

BUT IF WE CONSERVE ANGULAR MOMENTUM ABOUT BOTTOMMOST POINT THE TORQUE BY FRICTION BECOMES ZERO (AS FRICTIONAL FORCE PASSES THRU BOTTOMMOSTPOINT) & HENCE NOW ANGULAR MOMENTUM CAN BE CONSERVED

KARTHIK THE FORMULA WHICH YOU HAVE APPLIED MVR + IW IS ABOUT THE BOTTOMMOST POINT(NOT ABOUT THE CENTRE WHICH U THOUGHT

)

HOPE YOU UNDERSTOOD

CHEERS!!!!!!!!!!!!!!!!!!!!

karthik2007

Edit - I have conserved momentum about the center, as the I have taken MI to be 2mr^2/5. There is some other flaw. Are you sure that friction is an external force, and we cant include it in the system?

Edit #2 - I have done some crap in my drowsiness. Is the answer v/6?

priyesh

karthik u have conserved it aboutthe bottomost point and not the center

because

when taking i about bottommost point

angular momentum is I(about the bottommost point) * w

=(I(about com) + mr^2) * w (using parellel axis theorem )

= I (about com)w + mr^2w

and mr^2w is nothing but mvr

= I(about com)w + mvr (which is what u have done dude)

see H.C verma part 1 pg 191 q 28 alternative part

u'll understand what i 'm saying

shubham_sachdeva

hey priyesh.......when we will conserve friction about lowermost point then mom. should be equal to I(about lowermost point) . W + mvr.....but u hv written it as only IW....plzz clarify.....acc. to u a sphere whcih is rolling will hv momentum equal to IW not IW + mvr???

@karthik.... answer calculated above by u is 3v/7...it is right man.but ur method is not confirmed......

priyesh

hey shubham

angular momentum about lowermost point will be I(about lowermost point) * w

which is equal i(about com) * w + mvr

see my above post i've derived

that's why when karthik has used iw + mvr he is actually conserving momentum about lowermost oint so that friction doesn't come into the scene.

but he is getting confused because in the formula iw + mvr i is about centre but this can be written as (i + mr^2)*w which is ang. momentum about lowermost point

see ang. momentum about the centre will be simply I(about centre ) * w

and for ang. momentum about lowermost point = i(about centre)w + mvr

if u still have any doubts do ask

cheers!!!!!!!!!!!!!! &

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