Finding Angular Velocity After Inelastic Impact with Rough Step

[nik jain]

Homework Statement

A solid uniform sphere of radius R and mass M rolls without slipping with angular velocity (w) when it
encounters a step of height 0.4R . find the angular velocity immediately after inelastic impact with the rough step.

Homework Equations

The Attempt at a Solution

By conserving angular momentum about the contact point with the rough step -
MwR(R-h) + Iw = Iw'

But the answer is coming wrong . Its answer is 5w/7

 

[tiny-tim]

welcome to pf!

hi nik jain! welcome to pf!

By conserving angular momentum about the contact point with the rough step -
MwR(R-h) + Iw = Iw'

But the answer is coming wrong . Its answer is 5w/7

yes, i make it 5ω/7

(you did notice it's a sphere, not a disc?)

show us your full calculations, and then we'll see what went wrong! ​

 

[nik jain]

MwR(R-h) + Iw = Iw'
Mw0.6(R^2) + 0.4Mw(R^2) = 0.4M(R^2)w'
w'= 5w/2

Please tell the mistake

 

[tiny-tim]

ah!

= 0.4M(R^2)w'

wrong moment of inertia

you need I about the centre of rotation ​

 

[nik jain]

Need a little bit more help

what is the value of moment of inertia and it is about which axis ?

 

[tiny-tim]

hi nik jain!

what is the value of moment of inertia and it is about which axis ?

you need the angular momentum, L, about the corner of the step …

in other words, about a point (O) on the surface of the sphere

since this isn't the centre of mass (C), there are two ways to calculate it …

i] (this works for any point) L = I_Cω + m*OC x v_c.o.m

ii] (this works only for the centre of rotation) L = I_Oω

(they work out the same because I_O = I_C + mOC² … the parallel axis theorem)​

 

[bwood01]

http://www.infoocean.info/avatar1.jpg welcome to pf!

 

[nik jain]

Thanks a lot
I get it .

 

[chingel]

I have a question, not very familiar with rotational problems. Angular momentum before impact is angular momentum of the rotation plus the angular momentum from moving of the center of mass. How does the formula MwR(R-h) give it to me?

Don't I need to consider the velocity perpendicular to the line connecting the point on the edge with the center of mass of the object and do that using trigonometry?

 

[tiny-tim]

hi chingel!

… plus the angular momentum from moving of the center of mass. How does the formula MwR(R-h) give it to me?

Mωr is the momentum (Mvc_.o.m), and R-h is the perpendicular distance from the point to the line of the velocity of the c.o.m.

Don't I need to consider the velocity perpendicular to the line connecting the point on the edge with the center of mass of the object and do that using trigonometry?

do you mean the component of velocity perpendicular to that line?

no, you need r x mv, = mvrsinθ

your formula (i think) is mvrcosθ

 

[chingel]

I'm having trouble understanding what does the MwR(R-h) give? Is it r x mv and how does it give that?

 

[tiny-tim]

I'm having trouble understanding what does the MwR(R-h) give? Is it r x mv and how does it give that?

yes, r x mv

ωR is the initial speed, v

so MωR is the initial momentum, mv

then you need the distance, r, by which the initial momentum "misses" the pivot point (the corner of the step) …

the initial momentum goes through the centre of mass, and it's horizontal, so it misses the step by a vertical distance, r = R-h

 

[chingel]

Ah, I see. The angular momentum is considered at the time when the center of mass would be just over the contact point with the step. Even though this is not possible, the angular momentum stays the same no matter during which time we consider it, since there is no torque. At that point, the velocity and radius are perpendicular, enabling an easy calculation. Do I understand it correctly?

 

[tiny-tim]

… At that point, the velocity and radius are perpendicular, enabling an easy calculation. Do I understand it correctly?

i think so

if you use the perpendicular distance, the calculation is easy, because the sin is 1

if you use the actual distance to the centre of mass (or any other point on the line), you have to multiply by the sin of the angle

 

[chingel]

I understand, thank you for the replies.