A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with 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.
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It is given that Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, =10 π rad/s
Coefficient of kinetic friction, = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force.
From Newton’s 2nd law of motion,
we have, frictional force, f = ma
a = Acceleration produced in the objects
m = Mass
∴ a =
Use formula, v = u + at
= 0 + gt
= gt.........(1)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
Whrre, α is angular Acceleration
mgr = –Iα
∴ α = –mgr / I
Use formula,
= ω0 + (–mgr / I)t
Rolling starts when linear velocity, v = rω
∴ v = r (ω0– μkmgrt / I) .........(2)
Equating equations (1) and (2), we get:
gt = r (ω0– mgrt / I)
= rω0 – mgr²t / I ….(vi)
For the ring:
I = mr²
∴ μkgt = rω0 – μkmgr²t / mr²
= rω0 – μkgt
2μkgt = rω0
∴ t = rω0 / 2μkg
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80 s
For the disc: I = (1/2)mr2
∴ μkgt = rω0 – μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
3μkgt = rω0
∴ t = rω0 / 3μkg
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53 s
Since td > tr, the disc will start rolling before the ring.
Initial angular speed, =10 π rad/s
Coefficient of kinetic friction, = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force.
From Newton’s 2nd law of motion,
we have, frictional force, f = ma
a = Acceleration produced in the objects
m = Mass
∴ a =
Use formula, v = u + at
= 0 + gt
= gt.........(1)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
Whrre, α is angular Acceleration
mgr = –Iα
∴ α = –mgr / I
Use formula,
= ω0 + (–mgr / I)t
Rolling starts when linear velocity, v = rω
∴ v = r (ω0– μkmgrt / I) .........(2)
Equating equations (1) and (2), we get:
gt = r (ω0– mgrt / I)
= rω0 – mgr²t / I ….(vi)
For the ring:
I = mr²
∴ μkgt = rω0 – μkmgr²t / mr²
= rω0 – μkgt
2μkgt = rω0
∴ t = rω0 / 2μkg
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80 s
For the disc: I = (1/2)mr2
∴ μkgt = rω0 – μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
3μkgt = rω0
∴ t = rω0 / 3μkg
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53 s
Since td > tr, the disc will start rolling before the ring.
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1 answer · Physics
Best Answer
Pendulum period in seconds
T ≈ 2π√(L/g)
or, rearranging:
g ≈ 4π²L/T²
L ≈ T²g/4π²
L is length of pendulum in meters
g is gravitational acceleration = 9.8 m/s²
The pendulum length is not mentioned, so I assume you have to calculate it from the above.
period is 1 second.
L ≈ (1)²g/4π² = 0.248 m. That size expanding to 1.1 meters doesn't make sense.
I think you left part of the problem out, or copied it incorrectly.
hope it helps
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