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3 years ago

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A small metal block with mass of 140 g sits on a horizontally rotating turntable. The turntable makes exactly 2 revolutions each

second. The disk is located 6 cm from the axis of ration of the turntable. Answer each of the following question:
a. What is the frictional force acting on the disk?
b. The disk will slide off the turntable If it is located at a radius larger than 11 cm from the axis of rotation. What is the coefficient of static friction?

1 answer:

andre [41]3 years ago

3 0

Answer:

a) Frictional force acting on the disc = 1.326N

b) Coefficient of static friction = 1.77

Explanation:

mass of the metal block = 140 g = 0.14 kg

Frequency of the turn table = 2 Revs/sec

distance of the disc from the axis of the turn table, r = 6 cm = 0.06 m

a) Frictional force acting on the disc

F = ma

a = V²/r ; V = ωr;

a = ω²r²/r ; a = ω²r

F = mω²r

ω = 2πf = 2π * 2 = 4π

ω = 4π

F = 0.14 * (4π)² * 0.06

F = 1.326 N

b) Coefficient of static friction

F = μmg.............(1)

F = mω²r...........(2)

Equating (1) and (2)

μmg = mω²r

μ = ω²r/g

r = 11 cm = 0.11 m

g = 9.8 m/s²

μ = ( 4π)² * 0.11/9.8

μ = 1.77

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