what is the torque exerted on it? please PLEASE help. due tomorrow!!?

when the play butotn is pressed , a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. if the CD has a radius of 6.0 cm and a mass of 17 g. what is the torque exerted on it?

i know the answer is 0.0018Nm but i need to know how to work it out. so if u could PLEASE show me like all the steps u did that would be cool

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  • Dr D
    Lv 7
    1 decade ago
    Favorite Answer

    Torque = I*α

    I = 1/2 * m * r^2 = 1/2 * 0.017 * 0.06^2 kg m^2

    To find α we can adjust Newton's equaiton of motion to circular motion

    ω^2 = 0 + 2*α * θ

    (450*2π/60)^2 = 2 * α * 3*2π

    α = 58.9 rad/sec^2

    Torque = 1.80 x 10^-3 Nm

  • Anonymous
    1 decade ago

    I think I can help you.

    The torque required can be computed using the following equation.

    tau = alpha*Iz

    where tau is the torque

    alpha is the acceleration

    Iz is the moment of inertia for a cylinder

    Using reference 1, we see that

    Iz = 1/2*m*r^2 = 3.06*10^-5 *kg*m^2

    Where m = 18 grams

    r = 6 cm

    Alpha is the term that requires a bit of thought. I will work by analogy. If we were working with linear variables, we can say that

    vf^2 - vi^2 = 2*a*d

    Where vf is the final velocity of an object

    vi is the initial velocity of an object

    a is its acceleration

    d is the distance it moves.

    A similar relationship exists for angular velocity.

    wf^2 - wi^2= 2*alpha*theta

    Where wf is the final angular velocity (450*2*pi/sec)

    wi is the initial angular velocity (0 rad/sec)

    theta is the angle the object moves (3*2*pi)

    Solving for alpha, I get alpha = 58.905 rad/sec^2.

    Substituting into the initial equation, I get tau = 0.0018 N*m

    Source(s): 1. Moment of inertia for a cylinder http://en.wikipedia.org/wiki/List_of_moments_of_in...
  • gp4rts
    Lv 7
    1 decade ago

    The average angular velocity is 450/2 = 225 rev/min = 3.75 rev/sec; The no of rotations at this average is 225*t. Therefore the time it takes to make these revs is 3 = 3.75*t, or t = 0.08 sec. The angular acceleration is 7.5 (rev/sec)/.8 (sec) (It takes 0.8 sec to go from 0 to 7.5 rps). This is a = 9.375 rev/sec^2

    The formula for acceleration of a rotating body is

    T = I*a, were I is the moment of inertia. For a thin disc, the moment of inertia is I = M*r^2 / 2. Convert to std units M = 0.017 kg, and r = 0.06 m gives I = 3.06*10^-5 kg*m^2

    Convert angular acceleration to rad / sec^2. I rev = 2*π radians. This gives 9.375*2*π = 58.905 rad/sec^2

    Now compute the torque T = 58.905 * 3.06*10^-5 = 0.0018 N*m

  • Edward
    Lv 7
    1 decade ago

    A lot!

    T=I A

    I-moment of itertia

    A- (alpha) angular acceleration

    A=aT R

    aT- tangential acceleration

    R - radius

    at=( 2pi R rpm/60) 3 rev =( 2x 3.14....x 06 x 450/ 60)/ 3

    at= .94 m/sec^2

    A= aT/ R= .94 m/sec^2 /(.06 m)=15.7rad/sec^2

    I=0.5 m R^2= 0.5 x 0.017 x (0.6)^2=0.0000306kg m^2

    T= I A=0.0000306kg m^2 x 15.7 rad/sec^2

    T=0.0004794m N

    T= 480 E-6 mN

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  • 1 decade ago

    take acceleration =velo/sec=450/60 re/sxs

    torque=Moment of Inertia (angular acceleration )

    M.I for disc with refeence to tangent in the plane =moment of inertia with reference to doameter=MR(R)/4

    torque= (Mr x r)/4 X(a/r) here a/r is the angular acceleration

    substitute the values you will get the answer

    convert radius into meter and mass into kilogram

    bye- enjoy :Physics lecturer

    Source(s): Yahoo
  • 1 decade ago

    don't cheat!

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