Question

A grinding wheel, initially at rest, is rotated for 8 seconds with a constant angular acceleration of 5 rad/s^2. The wheel is then brought to rest, with a uniform negative angular acceleration, in 10 revolutions. (a) Determine the angular acceleration needed. (b) How lng did it take to bring it to rest? *

Answer 1

The velocity of the wheel after the 8 seconds will be

$\omega = \beta_1 \cdot t_1 = 5\,\mathrm{rad/s^2}\cdot 8\,\mathrm{s} = 40\,\mathrm{rad/s}.$

10 revolutions correspond to $L=10\cdot 2\pi = 20\pi\,\mathrm{rad}.$

This length is also equal to $L = \omega t_2 - \dfrac{\beta_2 t_2^2}{2}$ , where $\beta_2$ is the acceleration in question (here we write the conditions for modulus of acceleration and put minus sign).

Also $\omega - \beta_2 t_2$ should be equal to 0, because the wheel should stop. So we obtain two equations with two unknown values $(\beta_2 \,\mathrm{and}\, t_2)$ :

$20\pi = 40 t_2 - \dfrac{\beta_2 t_2^2}{2},\; \; 40 - \beta_2 t_2 = 0.$

We express $\beta_2$ from the second equation: $\beta_2 = \dfrac{40}{t_2}$ and substitute it into the first equation:

$20\pi = 40 t_2 - \dfrac{40 t_2}{2} , \\ 20\pi = 20t_2, \\ t_2 = \pi\approx 3.14\,\mathrm{s} \; \Rightarrow \; \beta_2 = \dfrac{40}{\pi}\approx 12.7\,\mathrm{rad/s^2}.$

The deceleration is $-12.7\,\mathrm{rad/s^2}.$