Rotational Kinematics HW Help?

A potter's wheel with a 35.9 cm radius rotates with a 2.91 rad/s2 angular acceleration. After 5.37 s, the wheel has rotated through an angle of 77.7 rad.

a)What linear distance did a point on the outer edge travel during the 5.37 s?

b)What was the initial angular velocity of the wheel?

c)What was the angular velocity of the wheel at 5.37 s?

d)What is the centripetal acceleration at 5.37 s?

2 Answers

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  • Mangal
    Lv 4
    4 weeks ago
    Favorite Answer

    Initial angular speed, ω₀ = (not known)

    Final angular speed, ω = (not known)

    Angular acceleration, α = 2.91 rad/s²

    Angle of rotation, θ = 77.7 rad

    Time duration, t = 5.37 s

    Using the formula θ = ω₀t + (1/2)αt²

    ... ... ... 77.7 = ω₀ (5.37) + (1/2) (2.91)(5.37)²

    ... ... ... ω₀ = (1/5.37) {77.7 - (1/2) (2.91)(5.37)²}

    => ... .. ω₀ = 6.66 rad/s [ans b]

    Using the formula ω = ω₀ + αt

    ... ... ... ω = 6.66 + 2.91x5.37

    => ... .. ω = 22.28 rad/s [ans c]

    Centripetal acceleration = ω²r ... ... ... [r = radius of circular motion (for the point)]

    ... ... ... = (6.66)² (0.359) ... ... ... [35.9 cm = 0.359 m]

    ... ... ... = 15.9 m/s² [ans d]

    Distance travelled by the point, s = ωr = 6.66 x 0.359 = 0.237 m (or 23.7 cm) [ans a]

  • oubaas
    Lv 7
    4 weeks ago

    77.7 = ωo*5.37+2.91/2*5.37^2

    ωo = (77.7-2.91/2*5.37^2)/5.37 = 6.656 rad/sec

    n = Θ / 2PI revs

    S = C*n = 2PI*r*Θ / 2PI = 0.359*77.7 = 27.89 m

    ω = ωo+α*t = 6.656+2.91*5.37 = 22.28 rad/sec

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