Torque on a rolling cylinder

[phalanx123]

I.ve been working on this question for two days now but still couldn't get the right answer. could someone help me please?

A uniform disc of mass m and rasius a, rotating at an angular speed ω, is placed on a flat horizontal surface. If the coefficient of friction is μ, find the frictional torque on the disc, and hence calculate the time it takes to com to rest.

Here is my work.

the frictional force on the disc is F=μN=μmg

therefore the frictional torque is T=Fr=μmga

since T also equals to Iα, where I is the moment of inertia and α is the angular acceleration.

Then α=T/I=(μmga)/I

using ω(i)=ω+αt where ω(i) is the final angular velocity and ω is the initial anglar velocity.

Then 0=ω+αt for it to stop

which gives t=ω/α=(ωI)/(μmga) since I for this disc is 1/2ma2

which then gives t=(ωa)/(2μg)

 

[siddharth]

Hi and welcome to PF phalanx123!

A uniform disc of mass m and rasius a, rotating at an angular speed ω, is placed on a flat horizontal surface. If the coefficient of friction is μ, find the frictional torque on the disc, and hence calculate the time it takes to com to rest.

Is your disc placed horizontally on the surface, or is it held vertically?

I'm guessing that the question means the disc is placed horizontally on the surface, while you've worked it out when the disc is kept vertically on the surface, and held in place.

 

[phalanx123]

Oh thanks siddharth, No wonder I got the wrong answer. I'll try to do it as placed horizontaly on the surface. Thanks again ^_^