jack asked in Science & MathematicsPhysics · 2 months ago

Physics Question HELPPP!!!?

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 5.0 revolutions.

If the CD has a radius of 6.5 cm and a mass of 20 g , what is the torque exerted on it?

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3 Answers

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  • oubaas
    Lv 7
    2 months ago

    angular speed ω = 2PI/60*n = 0.10472*450 = 47.12 rad/sec 

    acceleration angle Θ = 5*2PI = 31.42 radians 

    angular acceleration α = ω^2/2Θ = 47.12^2/62.84 = 35.34 rad/sec^2

    M.o.I. J = m/2*r^2 = 10*10^-3*6.5^2*10^-4 = 4.225*10^-5 kg*m^2

    torque T = J*α = 4.225*10^-5*35.34 = 1.5*10^-3 N*m (just 2 sign. digits)

  • 2 months ago

    The rotational inertia of a disk spinning on its own axis is (1/2)MR^2, so in this case it's

    (1/2)(0.020 kg)(0.065 m)^2 = 4.225 x 10^(-5) kg*m^2.

    The final angular speed is (450 * 2pi rad)/(60 s) = 47.12 rad/s, so 

    the angular acceleration is (47.12 rad/s)^2 / (2*5*2pi rad) = 35.34 rad/s^2.

    Torque = (I)(alpha)

    = (4.225 x 10^(-5) kg*m^2)(35.34 s^(-2))

    = 0.00149 N*m

  • 2 months ago

    450 rev/min x 1 min/60 s = 7.5 rev/s = ω

    ω = Δθ/Δt

    5 rev = 10π rad

    Δt = Δθ/ω = 10π/7.5 = 4.189 s

    α = Δω/Δt = 7.5/4.189 = 1.79 rad/s²

    τ = αI

    I = ½MR² = ½0.02(0.065)²

    τ = 1.79(½)0.02(0.065)² = 0.0000756 Nm

    I is moment of inertia in kg•m²

    I = cMR²

        M is mass (kg), R is radius (meters)

        c = 1 for a ring or hollow cylinder

        c = 2/5 solid sphere around a diameter

        c = 7/5 solid sphere around a tangent

        c = ⅔ hollow sphere around a diameter

        c = ½ solid cylinder or disk around its center

        c = 1/4 solid cylinder or disk around a diameter

        c = 1/12 rod around its center, R = length

        c = ⅓ for a rod around its end, R = length

        c = 1 for a point mass M at a distance R from       the axis of rotation

        c = 1/3 for a door, where r is the width

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