Normal force on slipping hemisphere

[Epiclightning]

Homework Statement

A hemispherical bowl of mass m and radius R is placed on a rough horizontal surface. Initially the centre C of the bowl is vertically above the point of contact with the ground(see figure).
Now the bowl is released from rest. Find the normal force acting on the hemisphere initially if the coefficient of static friction is 1/2 of that required to prevent initial slipping of the bowl.

Homework Equations

a =R(w), w= angular acceleration
Centre of mass of hollow hemisphere = R/2 distance from centre

Ma_x = nN (n = coeff. Of friction)
Mg - N = Ma_y

MgR/2 = 5/3MR^2 (w) ...Torque about lowermost pt
w = 3g/10R

The Attempt at a Solution

I already figured out the minimum coefficient of friction for no slipping to be 6/17. This matches with the solution. But I am unable to find the normal force corresponding to half of this value(3/17).
I have 2 equations relating horizontal and vertical accelerations of the CM to N and Mg (above) but this is not enough as I need another to eliminate variables.

 

[haruspex]

You have not used the fact that the slipping bowl maintains contact with the surface, and does not penetrate it.

 

[Epiclightning]

The vertical acceleration of the lowermost point should be 0, then at this instant?

 

[haruspex]

The vertical acceleration of the lowermost point should be 0, then at this instant?

Yes.

 

[Epiclightning]

But that would mean the normal is the same as before(no slipping) as

Vertical acceleration of lowest pt=
Accn.(CM) - component of angular acceleration
= (Mg-N)/M - (Rw)(2/√5) =0

This gives N = 17mg/20 as nothing has changed from when there was no slipping. So what am I doing wrong?

 

[Chestermiller]

Let's see your free body diagram.

 

[zwierz]

he vertical acceleration of the lowermost point should be 0, then at this instant?

this is absolutely not evident and it should be obtained from the equations of motion (if it is true indeed). Why do not you write the equations in vector form first?

 

[Epiclightning]

FBD attached

 

[zwierz]

MgR/2 = 5/3MR^2 (w) ...Torque about lowermost pt

I have doubts regarding this equation. These all should be carefully written from the theorems of dynamics

 

[Epiclightning]

Oh, I think i found a possible error - am I not allowed to apply torque = I*alpha about that point as it is a noninertial frame which is not the instantaneous axis of rotation or the centre of mass

 

[Epiclightning]

Ok yes that was the error. I solved it taking torque about the center of mass and got the correct value of alpha. Then I got the right answer(N = 85mg/118).
Thanks for the help

 

[zwierz]

am I not allowed to apply torque = I*alpha about that point as it is a noninertial frame which is not the instantaneous axis of rotation or the centre of mass

actually you are not allowed to use this formula about instantaneous axis of rotation also

 

[haruspex]

this is absolutely not evident and it should be obtained from the equations of motion (if it is true indeed). Why do not you write the equations in vector form first?

The normal force between two bodies is the minimum force normal to the plane of contact that prevents interpenetration. The point of contact is initially at rest. Therefore it neither accelerates downwards into the floor nor up away from the floor.

 

[haruspex]

actually you are not allowed to use this formula about instantaneous axis of rotation also

There are two standard points on a moving object that can be used as axes for this equation. One is the mass centre and the other is the instantaneous centre of rotation. (In fact, it turns out that the set of axis points for which the equation holds forms a circle passing through those two points, but that's not very useful since it is quite hard to calculate.)

 

[haruspex]

Ok yes that was the error. I solved it taking torque about the center of mass and got the correct value of alpha. Then I got the right answer(N = 85mg/118).
Thanks for the help

You can take moments about the lowest point provided you use the point of contact on the ground, i.e. a fixed point in space. The angular acceleration about that point includes a component from the linear acceleration of the mass centre.

 

[zwierz]

he normal force between two bodies is the minimum force normal to the plane of contact that prevents interpenetration. The point of contact is initially at rest. Therefore it neither accelerates downwards into the floor nor up away from the floor.

Another informal explanation. All assertions must follow from the equations of motion. I do not claim that it is wrong I just say that formulas are needed, not words.

There are two standard points on a moving object that can be used as axes for this equation. One is the mass centre and the other is the instantaneous centre of rotation.

The last part is wrong. General formulation of corresponding theorem for planar case is as follows. Let $A$ be a point of the rigid body and $S$ be its center of mass. Then
$$J_A\dot{\boldsymbol\omega}+m \boldsymbol{AS}\times \boldsymbol a_A=\boldsymbol M_A,\qquad (*)$$
here $J_A$ is the moment of inertia about an axis that is perpendicular to the plane and passes through the point $A$,
$m$ is a mass of the body
$\boldsymbol a_A$ is an acceleration of the point $A$,
$\boldsymbol M_A$ is a torque that the rigid body experiences about the point $A$ .
So from (*) it is clear that your assertion holds if only vectors $\boldsymbol a_A$ and $\boldsymbol{AS}$ are collinear.

 

[haruspex]

All assertions must follow from the equations of motion

Yet the equations themselves follow from observations about the physical circumstances. My argument in post #13 seems sufficiently rigorous to me.

General formulation of corresponding planar theorem is as follows.

We may be at cross purposes here. I agree that the contribution of the linear acceleration must be included. See my post #15. As long as you do that it is clearly valid to use the point in the body which is the instantaneous centre of rotation since that is effectively an inertial frame. What is perhaps surprising is that is not valid to use an arbitrary point fixed in the body, even if the linear contribution is included.

 

[zwierz]

We may be at cross purposes here. I agree that the contribution of the linear acceleration must be included. See my post #15.

Now I see post #15:

The angular acceleration about that point includes a component from the linear acceleration

Sorry, this phrase does not make sense: angular acceleration of a rigid body does not depend on points. I also do not understand how angular acceleration can include components of a linear acceleration. What does exactly this mean?

Anyway the second term in the left side of (*) has not disappeared yet.

 

[haruspex]

I also do not understand how angular acceleration can include components of a linear acceleration.

Consider a point mass accelerating along the x axis. If I use the point (0, 1) in the xy plane as my reference axis, the point mass has an angular acceleration about the reference axis.

See also section 5 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/