DLS
  • DLS
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
Physics
katieb
  • katieb
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DLS
  • DLS
@Mashy
DLS
  • DLS
@Mashy
yrelhan4
  • yrelhan4
@Mashy @shubhamsrg

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DLS
  • DLS
@Mashy
DLS
  • DLS
@yrelhan4 @RnR :P
yrelhan4
  • yrelhan4
:/
anonymous
  • anonymous
should we consider all the types of friction in question 1? :P.. we can only consider rotational rigth? :D..
DLS
  • DLS
you will require all 3,when you start the question you will realise :)
anonymous
  • anonymous
but at a time only one friction comes into play.. ! :-/
DLS
  • DLS
yes..start with static one
anonymous
  • anonymous
but it ll only act for a particular small amount of time.. which i dunno how we could calculate :P
DLS
  • DLS
:/ okay
anonymous
  • anonymous
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!
DLS
  • DLS
proceed with whatever u want :/
anonymous
  • anonymous
wait.. rolling friction wouldn't make it stop :D.. it should be the kinetic friction :D
DLS
  • DLS
arrre kar to :/
DLS
  • DLS
@Mashy ?
anonymous
  • anonymous
so as i was saying it should be rotational friction itself anyways wat about mass and the radius?!
DLS
  • DLS
M,R
anonymous
  • anonymous
lol just like tat you grant wishes huh? :D lol
DLS
  • DLS
assume anything duh
anonymous
  • anonymous
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!
DLS
  • DLS
in the end you will get F=-f
DLS
  • DLS
after applying torque equation n stuff
DLS
  • DLS
now?
anonymous
  • anonymous
what?!? f = (mur)N/R.... thats the formula for rolling friction right?!
DLS
  • DLS
/R? upon R?no simply u(r)N
anonymous
  • anonymous
it depends upon the dimensions of mur .. rolling friction can be expressed in both the ways :P
DLS
  • DLS
abe yr ata hai to kar kya ghante me reply kar rha h :/
yrelhan4
  • yrelhan4
^ Best response. :P
anonymous
  • anonymous
arre sorry.. i was wtching dbz abridged :D..
anonymous
  • anonymous
Can you draw a picture?
anonymous
  • anonymous
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.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
DLS
  • DLS
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!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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 \(v_1=\Large \frac {R\omega_0}{3}\)

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