Answer:
[tex]0.629\ \text{rad/s}^2[/tex] counterclockwise
[tex]9.98\ \text{s}[/tex]
Explanation:
[tex]r_1[/tex] = Small drive wheel radius = 2.2 cm
[tex]\alpha_1[/tex] = Angular acceleration of the small drive wheel = [tex]8\ \text{rad/s}^2[/tex]
[tex]r_2[/tex] = Radius of pottery wheel = 28 cm
[tex]\alpha_2[/tex] = Angular acceleration of pottery wheel
As the linear acceleration of the system is conserved we have
[tex]r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2[/tex]
The angular acceleration of the pottery wheel is [tex]0.629\ \text{rad/s}^2[/tex].
The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\omega_f[/tex] = Final angular velocity = [tex]60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}[/tex]
t = Time taken
From the kinematic equations of linear motion we have
[tex]\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}[/tex]
The time it takes the pottery wheel to reach the required speed is [tex]9.98\ \text{s}[/tex]
