Centripetal force and water in a Bucket

[Jimmy87]

Homework Statement

To explain why water falls out of a bucket when it is not being swung fast enough

Homework Equations

a = mv^2/r

The Attempt at a Solution

I have had a look on previous threads and nearly get it but still a bit confused. When the bucket is upside down and the required centripetal force is 9.81m/s^2 then the water will stay in the bucket and the tension in the string will be zero and the normal force from the bottom of the bucket on the water will also be zero. If the bucket was swung where the required centripetal acceleration was 5m/s^2 then gravity would provide too much and the water would fall out of the bucket. It's the description in my book that confuses me:

"In order for the water to stay in the bucket when it is upside down the water needs to be accelerated (pulled downwards) faster than what gravity would move the water". If water is being pulled downwards faster than gravity then wouldn't it fall out the bucket even faster not stay in the bucket? Also if the required centripetal acceleration is 5m/s^2 then surely the bucket would also fall so since they are both falling wouldn't the water still stay in the bucket until it hit the ground?

Any help is much appreciated!

 

[BvU]

If water is being pulled downwards faster than gravity then wouldn't it fall out the bucket even faster not stay in the bucket?

Can't make much of what this is trying to say. Anyway, there is no other downward pulling force.

Also if the required centripetal acceleration is 5m/s^2 then surely the bucket would also fall so since they are both falling wouldn't the water still stay in the bucket until it hit the ground?

Good point. And you are correct. So you don't just get the water on your head, but the bucket as well!

 

[Jimmy87]

Can't make much of what this is trying to say. Anyway, there is no other downward pulling force.

Good point. And you are correct. So you don't just get the water on your head, but the bucket as well!

Thanks. So if the bucket was being swung at 5m/s^2 then would the water and bucket fall at the same rate (i.e. the water would stay in the bucket the whole time until it fell on your head)?

 

[BvU]

At $\omega^2 r = 5$ m/s², gravity would win at an angle of 30 degrees above horizontal. The water pours out, rope goes slack, etc. No uniform circular motion possible (which, by the way is generally the case: g pulls down both at the bottom and at the top).

Wait for reasonable weather and do the experiment! Tip: use a small bucket.

 

[Andrew Mason]

Homework Statement

To explain why water falls out of a bucket when it is not being swung fast enough

Homework Equations

a = mv^2/r

The Attempt at a Solution

I have had a look on previous threads and nearly get it but still a bit confused. When the bucket is upside down and the required centripetal force is 9.81m/s^2 then the water will stay in the bucket and the tension in the string will be zero and the normal force from the bottom of the bucket on the water will also be zero. If the bucket was swung where the required centripetal acceleration was 5m/s^2 then gravity would provide too much and the water would fall out of the bucket. It's the description in my book that confuses me:

"In order for the water to stay in the bucket when it is upside down the water needs to be accelerated (pulled downwards) faster than what gravity would move the water". If water is being pulled downwards faster than gravity then wouldn't it fall out the bucket even faster not stay in the bucket? Also if the required centripetal acceleration is 5m/s^2 then surely the bucket would also fall so since they are both falling wouldn't the water still stay in the bucket until it hit the ground?

Any help is much appreciated!

You should give us the name of the physics text you are using and the page that this explanation appears on. Incorrect solutions in texts are not uncommon but they are often minor mistakes such as calculation errors. This is a conceptual error that indicates that the author has a fundamental misunderstanding of something rather basic. I would complain to your teacher.

AM