Q6.8 Torque and angular acceleration (S)

A pulley with moment of inertia 1.0×10-3 kg m2 about its axle, and a radius of 10 cm, is acted upon by a force applied tangentially at its rim. The magnitude of the force varies in time as

F=0.5t+0.3t2

where F is in newtons and t is in seconds. At t=3 s, what are

  1. its angular acceleration
  2. its angular velocity
Hide

Hint

Don’t forget how torque is defined - see section S5.6.
 

Solution

Reveal
Hide

Solution

Since the force acts tangentially at a distance r from the rotation axis, then τ=Fr. Using Newton’s second law in angular form (Key Point 5.19)
τ=Iα
we can calculate α and ω
  1. From the definition of torque and Newton’s 2nd law, we have
    \[ \alpha = \frac{\tau}{I} = \frac{rF}{I}=\frac{r}{I} \left( 0.5 t + 0.3t^2 \right) =420\,rad\,s^{-2}. \]
  2. To calculate ω we have to integrate (Key Point 5.6)
    \[ \omega(t) = \int_0^t \alpha dt = \int_0^t \frac{r}{I} \left( 0.5 t + 0.3t^2 \right) dt \]
    so that
    \[ \omega = \frac{r}{I}\left[ \frac{t^2}{4} + \frac{t^3}{10} \right]^3_0 = 495\,rad\,s^{-1} \]