Question

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The smalldrive-wheel has a radius of 2.20 cm and accelerates at the rate of 8.00 rad/s2, and it is in contact with the pottery wheel (radius 28.0 cm). Both wheels move without slipping.The rubber drive wheel rotates in the clockwise sense.
Required:
a. Find the angular acceleration (both magnitude and direction) of the large pottery wheel.
b. Calculate the tune it takes the pottery wheel to reach its required speed of 60 rpm. if both wheels start from rest.

Answer 1

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$.

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$