A particle moves 3.0 m along a circle of radius 1.5 m. (a) Through what angle does it rotate? (b) If the particle makes this trip in 1.0 s a

Question

A particle moves 3.0 m along a circle of radius 1.5 m. (a) Through what angle does it rotate? (b) If the particle makes this trip in 1.0 s at a constant speed, what is its angular velocity? (c) What is its acceleration?

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Phúc Điền 10 mins 2021-07-22T09:44:37+00:00 1 Answers 0 views 0

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    2021-07-22T09:45:43+00:00

    Answer:

    (a) 114.65° or 2.00 rad

    (b) 2.00 rad/s

    (c) 0 m/s²

    Explanation:

    (a)

    Note: As the particle moves along the circle, it forms an arc.

    Length of an arc = (Ф/360)×2πr

    L = (Ф/360)×2πr……………………. Equation 1

    Where L = distance at which the particle moves along the circle, Ф = angle of rotation of the particle, r = radius of the circle.

    make Ф the subject of the equation

    Ф = 360L/2πr…………………. Equation 2

    Given: L = 3 m, r = 1.5 m

    Substitute into equation 2

    Ф = 360(3)/(2×3.14×1.5)

    Ф = 1080/9.42

    Ф = 114.65° or 2.00 rad

    (b)

    Angular velocity(ω) = Ф/t

    ω = ΔФ/Δt…………………….. Equation 3

    Where ω = angular velocity, Δt = time taken to make the trip, ΔФ = angle through which the particle rotates

    Given: ΔФ = 2.00 rad, Δt = 1.0 s

    Substitute into equation 3

    ω = 2.00/1

    ω = 2.00 rad/s

    (c)

    α = dω/dt……………… Equation 4

    Where α = angular acceleration, dω = change in velocity, dt = change in time

    α = d(2.00)/dt

    Note: since ω is a constant, and the differentiation of a constant is equal to zero,

    therefore,

    α = 0 rad/s².

    But,

    a = αr……………. Equation 5

    Where a = acceleration of the particle

    Given: α = rad/s², r = 1.5 m

    Substitute into equation 5

    a = 0(1.5)

    a = 0 m/s²

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