## DLS Group Title 1)A disc has pure rotation.It has angular velocity w.It then comes into contact with a surface which has static friction us,kinetic friction uk and rolling friction ur,When will the disc stop? 2)A disc is placed on the ground.The co-efficient of friction is u,and the co-efficient of rolling friction is u_r.What is the minimum force required to move the disc?The force is applied at the bottom most point one year ago one year ago

1. DLS Group Title

@Mashy

2. DLS Group Title

@Mashy

3. yrelhan4 Group Title

@Mashy @shubhamsrg

4. DLS Group Title

@Mashy

5. DLS Group Title

@yrelhan4 @RnR :P

6. yrelhan4 Group Title

:/

7. Mashy Group Title

should we consider all the types of friction in question 1? :P.. we can only consider rotational rigth? :D..

8. DLS Group Title

you will require all 3,when you start the question you will realise :)

9. Mashy Group Title

but at a time only one friction comes into play.. ! :-/

10. DLS Group Title

11. Mashy Group Title

but it ll only act for a particular small amount of time.. which i dunno how we could calculate :P

12. DLS Group Title

:/ okay

13. Mashy Group Title

and besides i really don't think the static and kinetic have to be considered.. i mean how can you consider static? the wheel is already in rotation .. so no question of static.. and the moment it would come in contact it would start rolling OR slipping.. so only one would come into picture!

14. DLS Group Title

proceed with whatever u want :/

15. Mashy Group Title

wait.. rolling friction wouldn't make it stop :D.. it should be the kinetic friction :D

16. DLS Group Title

arrre kar to :/

17. DLS Group Title

@Mashy ?

18. Mashy Group Title

so as i was saying it should be rotational friction itself anyways wat about mass and the radius?!

19. DLS Group Title

M,R

20. Mashy Group Title

lol just like tat you grant wishes huh? :D lol

21. DLS Group Title

assume anything duh

22. Mashy Group Title

so now there would be a force acting creating a torque that would eventually sue the rolling.. so i guess all you have to do is finde the torque.. then you can find the deceleration!

23. DLS Group Title

in the end you will get F=-f

24. DLS Group Title

after applying torque equation n stuff

25. DLS Group Title

now?

26. Mashy Group Title

what?!? f = (mur)N/R.... thats the formula for rolling friction right?!

27. DLS Group Title

/R? upon R?no simply u(r)N

28. Mashy Group Title

it depends upon the dimensions of mur .. rolling friction can be expressed in both the ways :P

29. DLS Group Title

abe yr ata hai to kar kya ghante me reply kar rha h :/

30. yrelhan4 Group Title

^ Best response. :P

31. Mashy Group Title

arre sorry.. i was wtching dbz abridged :D..

32. wio Group Title

Can you draw a picture?

33. wio Group Title

When it comes to static and kinetic friction, my intuition, and it could be wrong, is that static friction is only meant as a thresh hold to find out whether the object will be moved. Kinetic friction on the other hand is something which applied a resting force over time.

34. Vincent-Lyon.Fr Group Title

What is your definition of µr (rolling friction coefficient)? It is important, because it is only if rolling friction exists that the disc will eventually come to a halt. Otherwise, it will roll for ever.

35. DLS Group Title

Your answer is right. I don't have the exact definition of µr (rolling friction coefficient) or even rolling friction. I just know that rolling friction is 1/100th of sliding friction and it depends on hardness of surface. Sorry,ill be glad to get some details about it!

36. Vincent-Lyon.Fr Group Title

The first phase of the motion is sliding. It is not so straightforward to find the time it will last. The answer is $$\tau_1=\Large \frac {R\omega_0}{3\mu _kg}$$ The remaining rotation velocity is $$\omega_1=\Large \frac {\omega_0}{3}$$ Velocity of centre of mass is $$v_1=\Large \frac {R\omega_0}{3}$$