Answer :
Answer
given,
diameter of compact disk = 12 cm = 0.12 m
angular velocity = 2000 rpm
[tex]\omega=2000\times \dfrac{2\pi}{60}[/tex]
ω = 209.43 rad/s
time taken to reach angular velocity = 3 s
moment of inertia of disk = 2.5 x 10⁻⁵ kg.m²
a) using equation,
ω = ω₀ + α t
209.43 = 0 + α x 3
α = 69.81 rad/s²
torque = I α
τ = 2.5 x 10⁻⁵ x 69.81
τ = 1.75 x 10⁻³ N m
b) using equation
[tex]\theta = \omega_0 t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta =0 \times t + \dfrac{1}{2}\times 69.81 \times 3^2[/tex]
θ = 314.145 rad
revolution = [tex]\dfrac{314.145}{2\pi}[/tex]
revolution = 50 revolutions
The amount of torque is applied to the disk is 1.75×10⁻³ Nm and the number of revolutions it makes before reaching full speed is 50 rpm.
What is torque?
The torque of a body is the rate of change of angular momentum of the body. Torque is the product of moment of inertia and the angular acceleration.
[tex]\tau=I\alpha[/tex]
Here, (I) is the moment of inertia and (α) is the angular acceleration.
The final angular velocity, of disk is 2000 rpm. In rad/s,
[tex]\omega=(2000)\dfrac{2\pi}{60}\\\omega=209.43\rm\; rad/s[/tex]
Angular acceleration of a body is the change of angular speed with respect to time. Thus,
[tex]\alpha=\dfrac{\Delta \omega}{t}\\\alpha=\dfrac{\omega-\omega_o}{t}[/tex]
Here, (ω₀) is the initial angular velocity and (t) is time.
- Part A) The amount of torque is applied to the disk-
Initial angular velocity of the disk is zero and final angular velocity is 209.43. The time taken is 3 seconds. Thus, the angular acceleration is,
[tex]\alpha=\dfrac{209.43-0}{3}\\\alpha=69.81\rm\; rad/s^2[/tex]
The disk's moment of inertia is 2.5x10^-5. Thus, the torque is,
[tex]\tau=2.5\times10^{-5}(69.81)\\\tau=1.75\times10^{-3}\rm\; Nm[/tex]
- Part B) Number of revolutions it make before reaching full speed-
By the equation of rotational motion,
[tex]\theta=\omega_ot+\dfrac{1}{2}\alpha t^2\\\theta=0+\dfrac{1}{2}(69.81)(3)^2\\\theta=314.145\rm \; rad\\\theta=\dfrac{314.145}{2\pi}\\\theta=50\rm\; rpm[/tex]
Thus, the amount of torque is applied to the disk is 1.75×10⁻³ Nm and the number of revolutions it makes before reaching full speed is 50 rpm.
Learn more about the torque here;
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