Inelastic collision in rotational motion

[zorro]

Homework Statement

Two blocks A and B are attached to a spring of force constant K and are placed on a horizontal surface. The coefficient of friction between block A and surface is zero while between block B and surface is μ. A spherical ball of mass M, radius R impinges on block A with angular velocity w and linear velocity u as shown.

The coefficient of restitution is 1/2 and coefficient of friction between the ball and block A is 1/√2. Find velocity of block B when spring suffers maximum deformation. Initially the spring is non deformed. (M = 1kg, Ma=3kg, u=10m/s, μ=0.2, Mb=10kg, k=300/16 N/m)

The Attempt at a Solution

The impulsive force due to friction provides A with some velocity.
Do we have to take the angular velocity of the ball into consideration for writing the impulse-momentum equation and Newton's Law of Restitution equation? The ball has an angular velocity wR not equal to u which makes things difficult. We also have 5 variables u',v,θ,w,R. Please throw some light on this.

 

[zorro]

anyone?

 

[tiny-tim]

Hi Abdul!

The impulsive force due to friction provides A with some velocity.
Do we have to take the angular velocity of the ball into consideration for writing the impulse-momentum equation and Newton's Law of Restitution equation?

There will also be an impulsive torque, equal to R times the impulsive friction force.

For collisions, you always have conservation of momentum, and you need one more equation for each component.

(Sometimes it's conservation of energy, but of course not here.)

In the y direction, it's the restitution equation, and in the x direction it's (ultimately) the angular momentum equation.

The ball has an angular velocity wR not equal to u which makes things difficult.

ωrR will be the relative speed.

 

[zorro]

I am writing the restitution equation along y-direction. Tell me if it is correct.

MwRcos45 + Mucos45 = 0.5(Mw'Rcosθ + Mu'cosθ), where θ is the angle made with the vertical after collision and u',w' are the respective velocities (In the upward direction)

 

[tiny-tim]

Abdul, please don't split your solution into bits and ask for each bit to be checked separately …

do the whole thing (or as much as you can), and then ask.

 

[zorro]

I was unsure (about assuming new angle with the vertical) of that equation.
Anyway this is the momentum conservation equation in x-direction-
musin45 + mwRsin45 = mv + mu'sinθ + mw'Rsinθ

Momentum will not be conserved along y-direction.

and in the x direction it's (ultimately) the angular momentum equation.

about which point?

In the end we have 3 equations and 4 variables.

 

[zorro]

do the whole thing (or as much as you can), and then ask.

I have done as much as I can!

 

[hikaru1221]

1. Do we have to take the angular velocity of the ball into consideration for writing the impulse-momentum equation?

Nope, if you meant it was linear momentum. Actually, the angular velocity is always taken into consideration, and it has to be so. BUT: it is finally canceled out. Here is the reason:

Consider a body (a system of mass points will do, not necessarily a rigid/ continuous body), whose COM is at point A. The total linear momentum of the body is the sum of linear momentums of all infinitesimal elements whose locations are conventionally points M_i:

$$\vec{p}_{total} = \sum m_i\vec{v}_i = \sum m_i(\vec{v}_{COM} + \vec{\omega} \times \vec{AM_i}) = \sum m_i\vec{v}_{COM} + \sum m_i(\vec{\omega} \times \vec{AM_i})$$

But: $$\sum m_i(\vec{\omega} \times \vec{AM_i}) = \vec{\omega} \times \sum m_i\vec{AM_i} = \vec{0}$$
as by definition of COM: $$\sum m_i\vec{AM_i} = \vec{0}$$

Therefore: $$\vec{p}_{total} = \sum m_i\vec{v}_{COM} = m\vec{v}_{COM}$$
where $$m = \sum m_i$$ is the total mass of the system.

So the conclusion is: for any system of mass points, the linear momentum of the whole system = "linear momentum" of COM; angular velocity doesn't explicitly appear, as it's already canceled out.

P.S.: The conservation of linear momentum applies to everything involved in the collision: sphere + mass A + mass B + spring + Earth, not just mass A + the point mass on sphere that touches mass A! However, spring is massless and thus has no momentum, mass B stays at rest during the collision, the Earth is not impacted horizontally (the 3rd point on the Earth needs a good reasoning behind it; find it yourself ), and so, for horizontal linear momentum, we are left with the sphere and mass A. The sphere = all the point masses contained in the sphere, not just the point of contact

2. Do we have to take the angular velocity of the ball into consideration for writing Newton's Law of Restitution equation?

I don't know Newton's law of restitution is not a universial law, and doesn't appear very frequently in scientific calculations (I'm not sure about engineering, e.g. mechanical/ material engineering). I doubt that there is a unified and official definition of the coefficient of restitution; or perhaps it's my shortcoming.
Check your textbook which is widely used in your country to see how they define it. The exams should apply the same or similar definition, so stick to that definition.3. How many unknowns to be determined for the collision?

4 actually. I see you listed R as an unknown. Those quantities like mass, length, initial speed, those that specify the intrinsic characteristics of the system & initial conditions, should be provided. (Classical) Mechanics problems are always deterministic provided that the system is well-specified and there are enough intial conditions.

If the problem doesn't provide the numerical value of R, then just leave R as R in the final result, after substituting all the values in

4. And how many equations to be written?

4 of course Those are:
1. The conservation of horizontal component of linear momentum of the sphere + mass A.
2. The equation for the rotation of the sphere. The change of linear momentum of the sphere is resulted from the frictional force F, and this F also exerts a change in angular momentum. Relate those two and eliminate F, you will arrive at one equation containing some of the 4 unknowns only, not F.
3. The equation for the vertical component of linear momentum of the sphere. This change is due to the normal force N. The frictional force F is related to N by the frictional coefficient between the sphere and mass A. Again, you will arrive at another equation containing some of the 4 unknowns only, not N or F.
4. The restitution equation.

 

[zorro]

Thanks for your reply,
some questions-
Is it right to assume that the angular velocity of the sphere after collision is w'=u'/R (the frictional force provides enough torque/impulse for that to happen)?
Is the different angle θ with the vertical due to the rotational motion of the sphere?

P.S.- The time allotted for this problem is 3 minutes. I don't want to get into the trouble of finding w',θ :(

 

[hikaru1221]

Is it right to assume that the angular velocity of the sphere after collision is w'=u'/R (the frictional force provides enough torque/impulse for that to happen)?

No, you can't assume that.

Is the different angle θ with the vertical due to the rotational motion of the sphere?

No and yes.
No because the angle is between the vertical axis & the linear velocity of the sphere, angular velocity not involved.
Yes because the value of angular velocity w of the sphere may affect the value of the linear velocity u'.

Good luck then. You will have to be either very fast, or finish other questions, each in less than 3 minutes, to have some extra time for this one.

 

[zorro]

@Hikaru-
after struggling for 2 hours...I found a very short method to solve this...tell me if it is right...

Downward and rightward directions are taken positive
m-mass of sphere
M-mass of block A
u' -velocity of sphere in the upward direction after collision
v- velocity of block A after collision in the rightward direction

1. Newton's Law of Restitution

0- (-u') = 0.5(u-0)
u'=0.5u (the linear velocity of point of contact w.r.t COM is horizontal, so any term containing w is not involved)

2. Impulse-momentum equation

For the sphere along vertical direction,
∫Ndt = m(u/2√2 + u/√2)
For the block along horizontal direction
∫μNdt = mv

solving for v we get v=u/4Now as soon as the block A gains this velocity, it starts to compress the spring. The block B doesnot move until the spring force is greater than 20N. It starts moving when the compression of the spring is 1.067m (app). How to find out the max. compression?

 

[hikaru1221]

The condition for max compression is that $$v_A = v_B$$, but when frictional gets involved, it seems that the only way is to solve everything formally, i.e. writing down Newton's 2nd law equations and solve for $$x_A$$ and $$x_B$$ and so on.

Is this a multiple choice question? If it is, there might be some way to make guesses based on the answers.

 

[zorro]

I solved it.
The minimum force required for B to move is 20N.
Max. energy provided by the block A = 0.5*3*(u/4)²
Max. compression possible =0.5*k*x²
Equating both we get x=1m.
so max. spring force = kx = 300/16=18.75N<20N
so max velocity=0

 

[zorro]

I have one problem :(
If I try to use momentum conservation along x-direction and impulse momentum theorem, I get wrong answer

1. Impulse momentum theorem along x-direction

∫μNdt=Mv
∫μNdt=m(Ux-U/√2), where Ux is the velocity of the COM of the ball along x-direction after collision

So Mv=m(Ux-u/√2)

Now applying the conservation of momentum along x-direction,

mu/√2 = Mv + mUx = m(Ux-u/√2) + mUx
So Ux=u/√2

Substituting this in any equation to find out v, we get v=0 !

What is the problem?