Disc
Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω0 =10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law
of motion, we have frictional force, f = ma μkmg= ma Where, a = Acceleration
produced in the objects m = Mass a = μkg … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μkgt
= μkgt … (ii)
The torque applied by the frictional force will act in a perpendicularly outward direction and cause a reduction in the initial angular speed.
Torque, τ= –Iα
α = Angular acceleration
μxmgr = –Iα
∴α=−μkmgrI ...(iii)
Using the first equation of rotational motion to obtain the final angular speed:
ω=ω0+αt=ω0+−μkmgrIt ...(iv)
Rolling starts when linear velocity, v = rω
∴v=r(ω0−μkgmrtI) ...(v)
Equating equations (ii) and (v), we get:
μkgt=r(ω0−μkgmrtI)=rω0−μkgmr2tI ...(vi)
For the ring: I=mr2∴μkgt=rω0−μkgmr2tmr2=rω0−μkgmtr
2μkgt=rω0∴tr=rω02μkg=0.1×10×3.142×0.2×9.8=0.80s ...(vii)
For the disc: I=12mr2∴μkgtd=rω0μkgmr2t12mr2=rω0−2μkgt
3μkgtd=rω0∴td=rω03μkg=0.1×10×3.143×0.2×9.8=0.53s ...(viii)
Since td > tr, the disc will start rolling before the ring.
