Force diagram of a spinning mass tied to a string

In summary, Mr. H does a demonstration with a bucket of water tied to a 1.3-meter long string. The bucket and water have a mass of 1.8 kg. Mr. H whirls the bucket in a vertical circle such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop.
  • #1
hndalama
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Homework Statement


In an effort to rev up his class, Mr. H does a demonstration with a bucket of water tied to a 1.3-meter long string. The bucket and water have a mass of 1.8 kg. Mr. H whirls the bucket in a vertical circle such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop.

Draw a free body diagram for the bucket for each location

Homework Equations

The Attempt at a Solution


At the top of the circle the the mass has gravity and tension going downwards, and at the bottom of the circle the mass has gravity downwards and tension upwards. but is there another force acting parallel to the tension and gravity? I think this would be the force that is causing the bucket to move in circular motion but I also think I might be thinking of a centrifugal force which is a fictitious force.
 
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  • #2
hndalama said:

The Attempt at a Solution


At the top of the circle the the mass has gravity and tension going downwards, and at the bottom of the circle the mass has gravity downwards and tension upwards.

Correct.

hndalama said:
...but is there another force acting parallel to the tension and gravity? I think this would be the force that is causing the bucket to move in circular motion but I also think I might be thinking of a centrifugal force which is a fictitious force.

Real forces must have a real physical cause. You have gravity and tension here. Where would another physical force come from?
 
  • #3
PeroK said:
Real forces must have a real physical cause. You have gravity and tension here. Where would another physical force come from?

right but at the top of the circle if there is only gravity and tension both acting downwards then why does the mass not drop straight down?
 
  • #4
hndalama said:
right but at the top of the circle if there is only gravity and tension both acting downwards then why does the mass not drop straight down?

Because it's moving! It accelerates directly down at this point, but it still has its horizontal speed.
 
  • #5
The simple view just has gravity and tension. Tension acting radially.

However in this case the angular velocity of the bucket isn't constant. Its faster at the bottom. So there is angular acceleration which means the tension force (probably) can't be radial all the time.

Its not clear which view the problem requires.
 
  • #6
PS Its likely they want the simple view but the other view explains how the bucket might be accelerated/decelerated as it rotates.
 
  • #7
CWatters said:
PS Its likely they want the simple view but the other view explains how the bucket might be accelerated/decelerated as it rotates.

I think you'll find that gravity explains the angular acceleration!
 
  • #8
PeroK said:
I think you'll find that gravity explains the angular acceleration!

Not when I did the sums but perhaps I made a mistake...

Energy in rotating mass = 0.5Iω2
Moment of inertia (for point mass on string) I = mr2
ω = v/r

Substituting for I and ω gives the familiar equation..
Energy = 0.5mv2

Energy at the bottom
= 0.5*1.8*6.42 = 36.9J

Energy at the top
= 0.5*1.8*3.92 = 13.7J

Difference = 36.9-13.7 = 23.2J

However the GPE = 1.8*9.81*2.6 = 45.9J

So the rope must be slowing it down on the way down and speeding it up on the way up. However I'm pretty sure this is over thinking the problem.
 
  • #9
CWatters said:
So the rope must be slowing it down on the way down and speeding it up on the way up. However I'm pretty sure this is over thinking the problem.

The difference in energy at the top and bottom is entirely due to the change in GPE. That's how you'd solve the problem.
 
  • #10
PeroK said:
The difference in energy at the top and bottom is entirely due to the change in GPE. That's how you'd solve the problem.
To find the difference in energy don't we have to account for the kinetic energy as well. so shouldn't it be KE(bottom) - KE(top) - PE(top) . PE at the bottom is 0.
so using CWatters numbers the change in energy is 36.9J -13.7J - 45.9J = -22.7 J

But how does considering the change in energy relate to whether tension is affecting the angular acceleration?
 
  • #11
hndalama said:
To find the difference in energy don't we have to account for the kinetic energy as well. so shouldn't it be KE(bottom) - KE(top) - PE(top) . PE at the bottom is 0.
so using CWatters numbers the change in energy is 36.9J -13.7J - 45.9J = -22.7 J

But how does considering the change in energy relate to whether tension is affecting the angular acceleration?

The waters are certainly muddied now. The change in energy of the bucket means the change in its KE. The change in its overall energy is 0, as the change in GPE is equal and opposite to its change in KE.

The assumption in this problem is that the string can sustain no lateral force, so when moving in a circle the tension is perpendicular to the velocity, hence does no work, hence neither adds nor subtracts from the overall energy. To get the object moving, the string would have to be pulled in a direction not perpendicular to its motion (i.e. not moved in a circle). I assume this problem assumes that steady circular motion has been reached.

Check the change in GPE and KE from the numbers given! My guess is they will be equal. If not, then maybe think again.
 
  • #12
hndalama said:

Homework Statement


In an effort to rev up his class, Mr. H does a demonstration with a bucket of water tied to a 1.3-meter long string. The bucket and water have a mass of 1.8 kg. Mr. H whirls the bucket in a vertical circle such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop.

Okay, whoever set this problem got the numbers wrong, I think. The gain in KE is almost exactly half the loss in PE, so either:

a) they forgot about the factor of ##1/2## for KE
b) they took the height difference to be ##1.3m## rather than ##2.6m##

Either way, the fact that one is precisely twice the other is a bit suspicious.
 
  • #13
PeroK said:
The change in its overall energy is 0, as the change in GPE is equal and opposite to its change in KE.

Check the change in GPE and KE from the numbers given! My guess is they will be equal. If not, then maybe think again.

this only applies if the mechanical/overall energy is conserved. the point I and I think @CWatters are making is that it isn't in this problem.
The calculation I made was to show that mechanical energy is lost. rather than the question being wrong can't we conclude that the loss of energy is due to air resistance?
 
  • #14
No I think I agree with perok in that they probably got the numbers wrong. Otherwise I think its a horrible problem.
 
  • #15
hndalama said:
this only applies if the mechanical/overall energy is conserved. the point I and I think @CWatters are making is that it isn't in this problem.
The calculation I made was to show that mechanical energy is lost. rather than the question being wrong can't we conclude that the loss of energy is due to air resistance?

If you assume the numbers in the question are correct, then you are wasting your time, in my opinion.
 
  • #16
hndalama said:
Mr. H whirls the bucket in a vertical circle such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop.
I think energy conservation is not the issue here. We can safely assume that Mr. H also does an unspecified amount of non-conservative work as he whirls the bucket. One only has to draw two snapshot FBDs at the top and bottom as the problem asks.
 
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Related to Force diagram of a spinning mass tied to a string

What is a force diagram of a spinning mass tied to a string?

A force diagram of a spinning mass tied to a string is a visual representation of the forces acting on a spinning mass that is connected to a string. It shows the direction and magnitude of each force, including the tension force from the string and the centripetal force keeping the mass in circular motion.

What is the purpose of creating a force diagram for a spinning mass tied to a string?

The purpose of creating a force diagram for a spinning mass tied to a string is to better understand and analyze the forces involved in the motion of the mass. This can help in predicting the behavior of the mass and determining the necessary conditions for maintaining circular motion.

What are the key components of a force diagram for a spinning mass tied to a string?

The key components of a force diagram for a spinning mass tied to a string are the mass, the string, and the forces acting on the mass. These forces include the tension force from the string, the weight force of the mass, and the centripetal force keeping the mass in circular motion.

How do you determine the direction of the forces in a force diagram for a spinning mass tied to a string?

The direction of the forces in a force diagram for a spinning mass tied to a string can be determined using Newton's laws of motion. The tension force from the string is always directed towards the center of the circle, the weight force is directed downwards, and the centripetal force is directed towards the center of the circle.

What is the relationship between the forces in a force diagram for a spinning mass tied to a string?

The relationship between the forces in a force diagram for a spinning mass tied to a string is that they must be balanced in order for the mass to maintain circular motion. This means that the sum of the forces in the horizontal and vertical directions must equal zero, and the centripetal force must be equal in magnitude to the tension force from the string.

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