Question

You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 13.2 cm. When the cylinder is rotating at 1.63 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

Answer 1

Answer:

centripetal acceleration$(a_{c})=13.8m/s^{2}$

Explanation:

convert 1.63rev/sec to rpm by multiplying by 60

= 1.63*60=97.8rpm

Convert this to rad/sec

1rpm =π/30 rad/sec

97.8rpm = 97.8 * (π/30 rad/sec)

              =10.25rad/sec

linear velocity=  angular velocity *radius

radius =13.2cm=13.2/100=0.132m

v=rω

v= 0.132*10.25

v=1.35m/s

centripetal acceleration = $\frac{v^{2} }{r}$

$a_{c}=\frac{1.35^{2} }{0.132}$

$a_{c}=13.8m/s^{2}$

Answer 2

Answer:

$a_{c}$ = 13.8 m/s².

Explanation:

The acceleration centripetal $a_{c}$ is given by:

$a_{c} = \frac{v^{2}}{r}$    (1)

where v: is the tangential speed and r: is the container radius

The tangential speed is equal to:

$v = \omega \cdot r$   (2)

where ω: is the angular velocity

Since 1 revolution is equal to 2π rad, the velocity (equation 2) is:

$v = 1.63 \frac{rev}{s} \cdot \frac{2\pi rad}{1rev} \cdot 0.132m = 1.35m/s$  

Now, by entering the velocity value calculated into equation (1) we can find the acceleration centripetal:

$a_{c} = \frac{(1.35m/s)^{2}}{0.132m} = 13.8m/s^{2}$

I hope it helps you!