IIT JEE , AIEEE Forum - Physics , Mechanics - Conceptual doubt on rotational mechanics.

Prakriteesh

We know that

=I

does not hold if axis frame is accelerating, except in a very special case (i.e. when the axis of rotation passes through the centre of mass).Does the relation L = Iw hold even when the axis frame is accelerating?

Does it hold in the special case mentioned above? Also, please check whether the reply I gave in http://www.goiit.com/posts/list/3926.htm is correct or not.

sss

what happens when your axis frame is accelerating???actually when your frame is acclerating,you have to apply a pseudo force on each and every particle,the resultant of which passes through the c.o.m of the body.so if you include the torque of that resultant force also,then you can definitely use the equation.actually,you can choose any axis ,you wii be able to solve the problem,only fact is that along with the torques of real forces,you have to apply the torque of that imaginary ,pseudo force also.if axis passes through c.o.m then torque due to that pseudo=0,bcoz it's point of application is the com itself.now,when comes to L,this formula is applicable only when you have chosen a axis at rest and unaccelerated wrt
ground i.e the axis must be fixed wrt ground (actually it is derived for that situation only)..but if you want to find Lusing any moving
axis,then you have to use L(vector)=L(along com)(vector)+m(r(vector)crossv(com)(vector))....actually i dont know how to type vectors ,therfore i have written in this way.

sss

when comes to your reply,you were correct.and one more thing,L=iw is derived for the situation when the axis is fixed relative to the ground.you can see the derivation.and the other one is when your axis frame is moving.you can check,whether i m right or not .take the axis at the pt. which is at rest and use the second eqn,you will automatically get the first eqn.

sss

thanks

Prakriteesh

Thanks sss. Just a few doubts left. You said that the resultant of the aplied force and the pseudo force passes through the c.o.m , but is that always necessarynecessary?. Imagine a situation where a tangential force acts on a single particle of a wheel and the axis frame is accelerating upwards. Then won't the resultant force be in an almost downward direction. dL/dt = T is valid in all frames. If the axis passes through the com, no pseudo torque is to be considered, even if the axis frame is accelerating. So, from the above relation dL/dt=T , won't L remain same if the axis passes through com, even if the axisframe is accelerating? Plz check the last message posted in http://www.goiit.com/posts/list/3481.htm.

sss

prakriteesh,that resultant pseudo force was the resultant of the pseudo forces acting on different particles of the rigid body.in this resultant,the applied,real forces are not considered. if you want to calculate the resultant of all real as well as imaginary(pseudo) forces,then their resultant can be anywhere.your frame is accelerating,so you have to apply pseudo force on every particle of the rigid body.i m talking of resultant of these forces.applied forces are not included here.and if you take axis at the centre which is accelerating,then no need to consider pseudo torques.actully for this reason only,when a rigid body is acclerating,the axis through com is preferred.isse pseudo torque lagaane kaa jhamela nahi rahtaa.but yes,if you take axis at other point,you can do question by applying pseudo torques.
when you take com frame no need to take pseudo torques then if torques of other applied forces=0,l will be conserved.

rao_prabhat

Hi Prakiteesh, I have understood your Question:
THE ANSWER IS THAT IT CANNOT BE APPLIED
Remember we cannot apply torque equation about a point, which is moving or accelerating because this would give the torque about the point to be zero. The above equation, which you have written for torque, applies for only about two points in a body. Those are:
1) Center of mass (if at rest)
2) Point which has zero velocity/acceleration
Therefore if a rigid body is placed in an accelerating axis frame, the pseudo forces arises from the condition makes all points on the rigid body to accelerate and the torque equation in the problem cannot be applied. Now coming to the question of applying L=I*omega here?s the explanation
L=I*omega is only the component of angular momentum due to rotation
There is also a component of L due to translation. In the given condition of the problem as I told u earlier every point experiences instantaneous acceleration and in the problem both rotation and translation happens for every point in the body certainly there are two components and L=I*omega cannot be solely be applied

--Hope this helps

Prakriteesh

I thank both of you, sss and rao_prabhat, for the help. Still, it will be nice if an expert clarifies the matter.

rao_prabhat

hey waht's wrong with my methid

Prakriteesh

Can't we use the equation after considering all the pseudo force torques just as I had done in the last msg of http://www.goiit.com/posts/list/3481.htm. ? Moreover, when we see a motion from the axis-frame itself, then we need to consider only the rotational part of L, won't we. Plz reply.

rao_prabhat

no you cannot calculate the torque due to pseudo forces because every body in the experiences instantaneous acceleration and the torqure about any point would be zero

Prakriteesh

Why is the torque zero when the particles are accelerating?

should depend only on rXF.Here r will not be changing as the axis itself is accelerating along with the particles. So, what is the reason for torque of being zero.

rao_prabhat

when you calculate the torque about a particle in the body which experiences acceleration it is zero because the force passes through the particle and the lever arm is zero

Prakriteesh

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