Question

A grinding disk (mass 2 kg and radius 0.10 m) is spinning at 800 rpm (revolutions per minute). You press a piece of metal against the edge of the disk and the disk slows to 740 rpm after a time of 10 seconds. Assume that the torque that you apply on the disk is constant.
a. What is 800 rpm in the unit rad/s?
b. What is 740 rpm in the unit rad/s?
c. What is the angular acceleration of the disk?
d. What is the moment of inertia of the disk?
e. How much torque do you apply on the disk?
f. How much impulse do you apply on the disk?
g. How much work do you do on the disk?

Answer 1

Answer

given,

mass of disk = 2 Kg

radius = 0.1 m

time = 10 sec

a) 800 rpm to rad/s

    = $800 \times \dfrac{2\pi}{60}$

    = $83.78\ rad/s$

b) 740 rpm to rad/s

    = $740 \times \dfrac{2\pi}{60}$

    = $77.49\ rad/s$

c) Angular acceleration

   $\alpha = \dfrac{\omega_f - \omega_i}{t}$

   $\alpha = \dfrac{83.78 -77.49}{10}$

   $\alpha =0.629\ rad/s^2$

d) moment of inertia of disc

   $I = \dfrac{1}{2}mr^2$

   $I = \dfrac{1}{2}\times 2 \times 0.1^2$

          I = 0.01 Kg.m²

e) τ = I α

   τ = 0.01 x 0.629

  τ = 6.29 x 10⁻³ N.m

f) impulse = Force x time

  $J= \dfrac{\tau}{r}\times t$

  $J= \dfrac{6.29 \times 10^{-3}}{0.1}\times 10$

          J = 0.629 Ns

g) Work done

      W = Δ KE

      $W = \dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

      $W = \dfrac{1}{2}\times 0.01\times (83.78^2-77.49^2)$

             W = 5.07 J