- #1
phalanx123
- 30
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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)
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)