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Over [174]
3 years ago
5

A turntable spins up from rest with an angular acceleration of 1.0 rad/s2. A rubber eraser of mass 20 g with coefficientof stati

c friction 1.0 is at a distance of 0.1 m from the center of the turntable, when it begins to spin. How long does ittake for the eraser to begin to slide off the turntable? Note that the tangential acceleration of the eraser is very small.
Physics
1 answer:
atroni [7]3 years ago
5 0

Answer:

9.89 seconds

Explanation:

Angular acceleration, α = 1 rad/s^2

mass of eraser, m = 20 g

μ = 1

r = 0.1 m

here, the centripetal force is balanced by the frictional force

m x r x ω^2 = μ m x g

r x ω^2 = μ g

ω^2 = 1 x 9.8 / 0.1 = 98

ω = 9.89 rad/s

Here, ωo = 0

Use

ω = ωo + α x t

9.89 = 0 + 1 x t

t = 9.89 seconds

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<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
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