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A disc rolls upon a straight line on a horizontal table, the flat surface of the disc being in contact with the table. If 
be the velocity of the center of the disc at 
, find the time after which the disc comes to rest. Given the coefficient of friction between the disc and the table is 
.
P.S: Let me add a diagram representing the situation:
Comments (19)
hello sir pls see if my soln is rite ...................... ASSUMPTION:normal force is uniform............................
IF WE DIVIDE THE DISC INTO VARIOUS ring discs OF MASS dm.....(let us assume seperate points of contacts crating a frictional force for each ring ..........................................
dm = [2(pi)(r)(dr)/(pi)(R)2]M ........................
NORMAL FORCE FOR EACH DISC...........
dmg...........................
torque about com for each disc .....................
[2(pi)(r)(dr)/(pi)(R)2]Mugr ....................(u = meu)..........
integrating the function from zero to R ...............we get net torque(say T) ...............
T/ I =A(ALPHA) ...........I = MR^2/2 .....................
0 = W - At ..................
W =At...........t=W /A ................................
pls correct me sir if im rong...................................
@shraman asa
your solution is not correct. The basic flaw is your assumtion that the force of friction is acting tangentially at each point of the elementary ring which is concentric with the disc. This would have been true had the disc been undergoing PURE rotation about an axis through the center and perpendicular to the plane. This, however, is not correct. Here the disc undergoes rotation about the central axis but it also posseses translatory motion. As a result of this, the resultant velocity of any point will not be tangential. Rather it will point in a direction that is along the vector sum of the translatory velocity and velocity accuired due to rotation about the central axis. So if you can modify slightly your approach, I think you would get the answer.
By the way, the required time is 
.
thats sin^2(theta) x rho!!!
The frictional force on each part of strip will be perpendicular to the shaded strip as it is rolling about the bottommost point. Taking x component of frictional force we get the net friction responsible for slowing it down. This force remains constant in magnitude. then use linear impulse = change in linear momentum. Sir is this approach correct??
hello sir!!!!!!!!
looking into a ring of mass m and RAD r (0<r<=R)breaking into elements dm.............................
dm = (rd@/2 pi r) m ......................(1) (taken theta as @)......
if u notice tan x = RWsin@ / rW + RWcos@ = Rsin@/r +Rcos@........................(2)
firction will act oppodite to direction of R(NOTE IN FUGIRE NOT RADIUS!!! ) vector.........
dT(TAU) = udmgcos(tan inv(rhs of eq 2) )r
dT = udmg(r +Rcos@ / root(R2 +r2 +2Rrcos@) )r.......................
put value of dm frm eqxn (1)................
integrate from 0 to 2pi..........................
we will get net torque for a ring of mass m and radius r..........................
but that will be a small element of the torque for the whole disc .........................
put m as 2 pi r dr / pi r2 M ............................
AND INTEGRATE THET FUNCTION AGAIN FROM 0 TO R.......................
ull get net torque ........... now using that find alpha and hence t...
sir pls tell me if my approach is correct.............
There is still some trouble. To be specific, you must find a relation between r and theta. Supposing that you do, it is still hopeless to perform the integration you have obtained.
There is, however, a better way if you take the elementary rings centered at the instanatneous axis of rotation (in this case they wont be complete rings; they will be arcs actually). Care to try?
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