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2 May, 14:23

A copper block rests 38.9 cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is 0.40. The turntable starts from rest and rotates with a constant angular acceleration of 0.60 rad/s2. After what time interval will the block start to slip on the turntable

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  1. 2 May, 14:40
    0
    t = 5.291 s

    Explanation:

    The copper block starts moving when the force responsible for the circular motion matches the frictional force.

    The frictional force is given by μN = μ mg where N = Normal reaction on the block.

    μ = coefficient of static friction = 0.4

    The force causing circular motion is given by mv²/r = mrw²

    mrw² = 0.4 mg

    rw² = 0 4g

    r = 38.9 cm = 0.389 m

    g = 9.8 m/s²

    w = √ (0.4*9.8/0.389)

    w = 3.174 rad/s

    We then use equation of motion for the turntable to find out when its angular velocity reaches 3.174 rad/s which is enough to move the block

    w = w₀ + αt

    w = 3.174 rad/s

    w₀ = 0 rad/s (the turntable was initially at rest)

    α = 0.60 rad/s²

    t = ?

    3.174 = 0 + 0.6t

    t = 5.291 s
  2. 2 May, 14:51
    0
    Fmax = μFn

    Where,

    Fn = normal force

    = mg

    μ = coefficient of static friction

    Fmax = 0.4mg

    Angular force, Fi = mω^2r

    mω^2 * r = 0.4mg

    ω^2 * r = 0.4g

    ω = sqrt (0.4 * 9.81/0.389)

    = 3.18 rad/s

    Using equations of angular motion,

    ωf = ωi + αt

    ωi = 0 rads

    t = ωf/α

    = 3.18/0.6

    = 5.29 s.
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