IIT JEE , AIEEE Forum - Physics , Mechanics

elastiboysai

A Uniform disc of mass m and radius r is rotated about an axis passing through

its centre and perpendicular to its plane with angular velocity

.It is placed on a rough horizontal plane with the axis of the disc vertical.

Coeff of friction between disc and surface is

.

Find time when disc stops rotating and

angle rotated by the disc before stopping.

akhil_o

let friction=u

torque=mg*u=mgu

alpha=mgu/I

=1/2gu/r*r

so we have

w²=2*alpha*theta

or angle rotated= 2r²w²/gu

thetea= wt+1/2alpha(t*t)

substitute value of theta and find time taken

its coming out to be

t=2r²w(

3 - 1)/ug

elastiboysai

Hi akhil,

did u get the qn rightly?

the axis is vertical -disc is placed horiz.

the answers are

3Wr/4

g

and3W^2r/8

g.

karthik2007

Are you sure about the answer? I got t = wR/4ug

(frictional torque = 2umgR)

elastiboysai

hi karthik,

the ans i have gvn is wat was gvn in the buk. 3Wr/4ug

but arihant books are known to be wrong.

can u give ur full proof .??

thanx

karthik2007

Here is my method.

Here there is no direct way of finding the frictional torque, so consider an annular ring. Its area = 2

rdr.

mass = m/

r² x m' = 2mdr/R.

Normal rxn = 2mgdr/R, so force due to friction = 2umgdr/R.

Frictional torque about center = 2umgdr x r

Hence, torque for the disc = _{[0 ]}

^{[r ]} 2u(2pir)rgdr = 2uMgr/3

Hence we get the desired result.

karthik2007

Well???

elastiboysai

let o=sigma,

dN=dm g

=o 2pirdrg

dF=udN

dT =urdN=uro2pirdrg

integr

T= o2piug*r^3/3

im gettin frictional torque as 2umgr/3

wid dis im gettin d desired result.

I think in the first step u have mixed up radius of disc and rad. of elemental ring.

elastiboysai

let o=sigma,

dN=dm g

=o 2pirdrg

dF=udN

dT =urdN=uro2pirdrg

integr

T= o2piug*r^3/3

im gettin frictional torque as 2umgr/3

wid dis im gettin d desired result.

I think in d 1st step ur cancellin is wrong as one is rad of disc n oder radius of elemental ring

elastiboysai

sorry for 2 posts. browser problem

abhijeet_0201

the frictional torque =2umgr/3 not 2/3umg
maybe a typing mistake by elastiboysai!!

the method is absolutely correct

both the answers r coming using simple eq of rotational motion.........so whats the prob????????

anchitsaini

T=Torque
sigma=s=M/pie r^2
dT=udmgr
=us2pierdrgr
=2uspie rsquare dr g
T=2uMgr\3
=Ialpha
alpha=4uMg/3r
w=alpha t
t=3wr/4ug

karthik2007

Oops.... sorry. Lol, I cancelled the radii, WHICH IS WRONG. The small r is the radius of the elemental ring. I did the calculations on the comp, so I guess made a flaw there. Thanks for pointing that elastiboysai. However, the method is correct.

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