Question

Professor Stefanovic is spinning a bucket of water by extending his arm and rotating his shoulder in class to show the effects of angular acceleration. When the bucket is upside down (exactly above his head) the water does not fall out. The mass of the water in the bucket is 2kg. The rope that is attached to the bucket is lm long, and his arm is 75 cm long.
1. What is the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out? 2. If he detaches the rope and spins the bucket by holding it with his hand, what is the minimum required angular velocity in order for the water not to fall out?

Answer 1

Answer:

a ) 2.368 rad/s

b) 3.617 rad/s

Explanation:

the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out can be determined by applying force equation in a circular path

i.e

$F_{inward } = F_G + F_T$   ------ equation (1)

where;

$F_{inward} = m *a_c$

$F_{inward} = m*r* \omega^2$

Also

$F_G = m*g$

$F_T = 0$          since; that is the initial minimum angular velocity to keep the water in the bucket

Now; we can rewrite our equation as :

$mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \ } ------ equation \ \ \ {2}$

So; Given that:

The rope that is attached to the bucket is lm long  and his arm is 75 cm long.

we have our radius r = 1 m +  75 cm

= ( 1 + 0.75 ) m

= 1.75 m

g = acceleration due to gravity = 9.81 m/s²

Replacing our values into equation (2) ; we have:

$\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \ rad/s$

b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m

∴

$\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \ rad/s$