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
- its angular acceleration
- its angular velocity
Solution
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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
- From the definition of torque and Newton’s 2nd law, we have
![\[ \alpha = \frac{\tau}{I} = \frac{rF}{I}=\frac{r}{I} \left( 0.5 t + 0.3t^2 \right) =420\,rad\,s^{-2}. \]](mastermathpng-0.png)
- To calculate ω we have to integrate (Key Point 5.6)
so that
![\[ \omega(t) = \int_0^t \alpha dt = \int_0^t \frac{r}{I} \left( 0.5 t + 0.3t^2 \right) dt \]](mastermathpng-1.png)
![\[ \omega = \frac{r}{I}\left[ \frac{t^2}{4} + \frac{t^3}{10} \right]^3_0 = 495\,rad\,s^{-1} \]](mastermathpng-2.png)