Homework Help: A question on rotation

1. Mar 20, 2012

nik jain

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
By conserving angular momentum about the contact point with the rough step -
MwR(R-h) + Iw = Iw'

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2. Mar 20, 2012

tiny-tim

welcome to pf!

hi nik jain! welcome to pf!
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!

3. Mar 20, 2012

nik jain

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

4. Mar 20, 2012

tiny-tim

ah!
wrong moment of inertia

you need I about the centre of rotation

5. Mar 20, 2012

nik jain

Need a little bit more help

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

6. Mar 20, 2012

tiny-tim

hi nik jain!
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 = ICω + m*OC x vc.o.m

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

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

7. Mar 20, 2012

bwood01

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

Last edited by a moderator: May 5, 2017
8. Mar 21, 2012

nik jain

Thanks a lot
I get it .

9. Mar 21, 2012

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?

10. Mar 21, 2012

tiny-tim

hi chingel!
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.
do you mean the component of velocity perpendicular to that line?

no, you need r x mv, = mvrsinθ

your formula (i think) is mvrcosθ

11. Mar 21, 2012

chingel

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

12. Mar 21, 2012

tiny-tim

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

13. Mar 21, 2012

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?

14. Mar 21, 2012

tiny-tim

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

15. Mar 21, 2012

chingel

I understand, thank you for the replies.