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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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\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

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