Question

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 12 cm. When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?a. 4.8 m/s²b. 19 m/s²c. 27 m/s²d. 9.8 m/s²e. 6.0 m/s²

Answer 1

Answer:

Option B

$a=19 m/s^{2}$

Explanation:

Given information

Radius of container, r=12cm=12/100=0.12m

Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π

Angular velocity, $\omega=2*2\pi =12.56637061$

We know that speed, $v=r\omega$

Centripetal acceleration, $a=\frac {v^{2}}{r}$ and substituting $v=r\omega$ we obtain that

$a=r\omega^{2}$

Substituting \omega for 12.56637061  and r for 0.12

$a=0.12*(12.56637061)^{2}=18.94964045 m/s^{2}$

Rounded off, $a=19 m/s^{2}$