Physics Question about Rotational Dynamics

[leoni101]

---

Members do not see this ad.

I am getting a different answer than than what this question's solution is saying. Could any of you guys help me see what I'm doing wrong?

A child spins a bucket in a vertical circle with a radius of 65 cm. The weight of the contents seems to change as the bucket rises and falls along the circular path. How fast must the bucket travel at the top of the circle so the contents seem to weigh 2 times their weight?

The way I see this, if you draw a FBD, for the force (which is weight) to be two times normal, the sum of the forces need to be two times as when it is standing still. So the forces acting on the bucket when it is at the top of the circle will be mg + m(v^2)/r = 2mg. This can be written as m(v^2)/r = mg.

I come out with 2.52 m/s but the answer says 4.37 m/s. What am I doing wrong?

 

[Pisiform]

I am getting a different answer than than what this question's solution is saying. Could any of you guys help me see what I'm doing wrong?

A child spins a bucket in a vertical circle with a radius of 65 cm. The weight of the contents seems to change as the bucket rises and falls along the circular path. How fast must the bucket travel at the top of the circle so the contents seem to weigh 2 times their weight?

The way I see this, if you draw a FBD, for the force (which is weight) to be two times normal, the sum of the forces need to be two times as when it is standing still. So the forces acting on the bucket when it is at the top of the circle will be mg + m(v^2)/r = 2mg. This can be written as m(v^2)/r = mg.

I come out with 2.52 m/s but the answer says 4.37 m/s. What am I doing wrong?

This is a very basic question. Think about it in this way. If you release an inverted bucket full of water, the water does not fall off the bucket right. Why? Because both the bucket and the water inside are accelerating downward are 9.8m/s^2.

Now, if you pull the inverted bucket down simultaneously as you release it, there would even be more chance that water won't fall. This is because now bucket is falling at an acceleration greater than 9.8m/s^2 (9.8 + acceleration with which you are pulling) while water is still falling at only 9.8m/s^2. This would make it much harder for water to fall off.

However, if you release the inverted bucket but some how slows it down, its acceleration would be less than 9.8m/s^2 while the water is still falling at 9.8m/s^2. This would cause the water to fall off the bucket.

Now look at the scenario with bucket at the top as the question is asking. At the top we know these forces are acting on the bucket: Centripetal Force, Force due to gravity and Normal force. Force due to gravity is acting downwards and want to make water fall at 9.8m/s^2. Also, Normal force is acting down. Both of these forces are working together to make water fall faster than the bucket but water is not falling right ... why? Because the centripetal force is preventing this to happen:

So,

Fc = Fg + N

mV^2/r = mg + ma (m cancels out)

v^2/r = g + a (they say it should feel two times its weight - means a force of 2g)

V^2 / 0.65 = (2g) + a

v^2 / 0.65 = (2*9.81) + 9.8

v = 4.37m/s

 

[leoni101]

This is a very basic question. Think about it in this way. If you release an inverted bucket full of water, the water does not fall off the bucket right. Why? Because both the bucket and the water inside are accelerating downward are 9.8m/s^2.

Now, if you pull the inverted bucket down simultaneously as you release it, there would even be more chance that water won't fall. This is because now bucket is falling at an acceleration greater than 9.8m/s^2 (9.8 + acceleration with which you are pulling) while water is still falling at only 9.8m/s^2. This would make it much harder for water to fall off.

However, if you release the inverted bucket but some how slows it down, its acceleration would be less than 9.8m/s^2 while the water is still falling at 9.8m/s^2. This would cause the water to fall off the bucket.

Now look at the scenario with bucket at the top as the question is asking. At the top we know these forces are acting on the bucket: Centripetal Force, Force due to gravity and Normal force. Force due to gravity is acting downwards and want to make water fall at 9.8m/s^2. Also, Normal force is acting down. Both of these forces are working together to make water fall faster than the bucket but water is not falling right ... why? Because the centripetal force is preventing this to happen:

So,

Fc = Fg + N

mV^2/r = mg + ma (m cancels out)

v^2/r = g + a (they say it should feel two times its weight - means a force of 2g)

V^2 / 0.65 = (2g) + a

v^2 / 0.65 = (2*9.81) + 9.8

v = 4.37m/s

Thanks. This certainly helps quite a bit. But how do you come up with 9.8 as 'a' in 'ma' for normal force when it sits at the top of the loop? F=ma I know, but why use 9.8? Is that the acceleration of the bucket against the contents? Sorry for asking such a basic question. I've been so focused on ochem 2, physics 2, and biology sections I've forgotten so much of the physics 1 stuff.

 

[Pisiform]

'a' is the acceleration due to gravity. In this case it's normal force is equal to weight but not 'weight' 'weight' since the actual weight has 2gs but weight under normal situation. You can call it mg for your convenience.

 

[leoni101]

'a' is the acceleration due to gravity. In this case it's normal force is equal to weight but not 'weight' 'weight' since the actual weight has 2gs but weight under normal situation. You can call it mg for your convenience.

Ok, I think I got it. Thanks!

 

[Ssina]

yeah the question is just pointing out the fact (which Pisiform explained) that in the FBD at the bottom of the circle the weight is pointing downward, while the normal force is upwards so Fc=ma-mg, which if you rearrange will give you the formula you wrote down (mg+ mv^2/r=mg)

but at the top, relating to the question, both are pointing downwards toward the center of the circle, hence the calculations Pisiform did...