A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with an initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is µk = 0.2.
Answers
Answer:
Radii of the ring and the disc, r = 10 cm = 0.10 m
Initial angular speed, ω0 =10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
The motion of the two objects is caused by the force of friction. According to Newton’s second, the force of friction, f = ma
μkmg= ma
Where,
a = Acceleration produced in the disc and the ring
m = Mass
∴ a = μkg . . . . . .. . . . . . . ( 1 )
Using the first equation of motion :
v = u + at
= 0 + μkgt
= μkgt . . . . . . . . . . . . . . ( 2 )
The frictional force applies a torque in perpendicularly outward direction and reduces the initial angular speed.
Torque, T= –Iα
Where, α = Angular acceleration
μkmgr = –Iα
∴ α = -μkmgr / I . . . . . . . . . ( 3 )
According to the first equation of rotational motion, we have :
ω = ω0 + αt
= ω0 + (-μkmgr / I )t . . . . . . . .( 4 )
Rolling starts when linear velocity, v = rω
∴ v = r (ω0 – μkmgrt / I ) …(5)
Using equation ( 2 ) and equation ( 5 ), we have:
μkgt = r (ω0 – μkmgrt / I )
= rω0 – μkmgr2t / I . . . . . . . . ( 6 )
For the ring:
I = mr2
∴ μkgt = rω0 – μkmgr2t / mr2
= rω0 – μkgt
2μkgt = rω0
∴ t = rω0 / 2μkg
= ( 0.1 × 10 × 3.14) / (2 × 0.2 × 10 ) = 0.80 s . . . . ( 7 )
For the disc: I = (1/2)mr2
∴ μkgt = rω0 – μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
3μkgt = rω0
∴ t = rω0 / 3μkg
= ( 0.1 x 10 × 3.14) / (3 × 0.2 × 9.8 ) = 0.53 s …..( 8)
Since tD > tR, the disc will start rolling before the ring.
Explanation:
Explanation:
simultaneously, with an initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic fric