What is the meaning of work done for non-uniform circular motion?

Click For Summary
SUMMARY

The discussion centers on the concept of work done in non-uniform circular motion, specifically addressing the relationship between work, kinetic energy (KE), and potential energy (U). The equation for work is clarified as W = F x, where F is the force and x is the distance over which it acts. The participants emphasize the distinction between conservative and non-conservative forces, noting that in non-uniform circular motion, the work done is expressed as ΔKE = W - ΔU. This indicates that both work and changes in potential energy contribute to changes in kinetic energy.

PREREQUISITES
  • Understanding of non-uniform circular motion
  • Familiarity with the work-energy theorem
  • Knowledge of kinetic and potential energy concepts
  • Ability to manipulate equations involving force and distance
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn about conservative vs. non-conservative forces
  • Explore examples of non-uniform circular motion in physics
  • Review potential energy calculations in various contexts
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of circular motion and the principles of work and energy in mechanical systems.

hhjjy
Messages
13
Reaction score
6
Homework Statement
In her hand a softball pitcher swings a ball of mass 0.250 kg around a vertical circular path of radius 60.0 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.0 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 15.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?
Relevant Equations
$$ \sum {W_{other forces}} = W = \Delta K + \Delta U + \Delta E_{int} $$
$$ E_{mech} = \Delta K + \Delta U $$
This is my solution ,and I just use the definition .But I still feel unclear about the concept of non-conservative force.$$ W = F x = 30N (\frac{1}{2}\pi r ) = 56.2 J $$
$$ E_{system} = \Delta K + \Delta U = W $$
$$ (K_{f}- K(i))+(U(f)-U(i)) = W $$
$$ (\frac{1}{2} *m{V_{f}}^2 -\frac{1}{2}*m{V_{i}}^2)+(mgh_{f}-mgh{i}) = W) $$
$$ V_{f} = \sqrt{V_{i}^2 + 2gh_{i} + \frac{2W}{m}} $$

I want to ask
1. Whether my solution is wrong.
2. What does the meaning of works done on a non-uniform circular motion?
 
Last edited:
  • Like
Likes   Reactions: PhDeezNutz
Physics news on Phys.org
I'm not sure I understand your solution. In this case, the work done means the increase in kinetic energy caused by the force applied in the direction of motion.
 
Can you explain that what 's the different between $$ E_{system} = \Delta K + \Delta U = W_{circular} $$ and $$ E_{system} = \Delta K + \Delta U = -W_{circular} $$ .And thank you for your reply .
 
Last edited:
hhjjy said:
Can you explain that what 's the different between $$ E_{system} = \Delta K + \Delta U = W_{circular} $$ and $$ E_{system} = \Delta K + \Delta U = -W_{circular} $$ .And thank you for your reply .
Work is the scalar product of force and direction of motion. If they are in the same direction, then work is positive, otherwise it's negative.

Also, if work is positive then KE is increasing, so the correct relationship must be $$\Delta KE = W - \Delta U$$ That's the way I remember it.
 
PeroK said:
Work is the scalar product of force and direction of motion. If they are in the same direction, then work is positive, otherwise it's negative.

Also, if work is positive then KE is increasing, so the correct relationship must be $$\Delta KE = W - \Delta U$$ That's the way I remember it.
So If we take another examples ,such as uniform circular motion or a pen fell from sky, $$ \Delta KE = 0 - \Delta U $$ ,which means there is no force working on it ,so we call it conservative of energy .

Non-uniform circular motion is $$ \Delta KE = W - \Delta U $$, which means there is other force work on it,so we call it non-conservative of energy .
 
Last edited:
  1. Please post workings in ## \LaTeX ## not in an image.
  2. You haven't posted a solution (the question asks for a speed).
  3. The first statement in your workings ## W = \displaystyle {\int} F x ## is not correct. You can write ## W = F x ##, ## W = \displaystyle \int F \ dx ##, or even ## W = \displaystyle \int F ##, but ## \displaystyle \int F x \ dx = \frac {F x^2} 2 ## which is not correct.
  4. The question states "the speed of the ball at the top of the circle is..." and asks for the speed at the bottom of the circle. Over what distance does the force of 30.0N act?
  5. Is the force from the pitcher's arm the only force acting on the ball?
 
pbuk said:
  1. Please post workings in ## \LaTeX ## not in an image.
  2. You haven't posted a solution (the question asks for a speed).
  3. The first statement in your workings ## W = \displaystyle {\int} F x ## is not correct. You can write ## W = F x ##, ## W = \displaystyle \int F \ dx ##, or even ## W = \displaystyle \int F ##, but ## \displaystyle \int F x \ dx = \frac {F x^2} 2 ## which is not correct.
  4. The question states "the speed of the ball at the top of the circle is..." and asks for the speed at the bottom of the circle. Over what distance does the force of 30.0N act?
  5. Is the force from the pitcher's arm the only force acting on the ball?
Thank you . I have corrected it .
4. It works along half of the circle .
5. I am not sure , but I think the pitcher's arm is the only force on the ball. I am confuse that whether gravity has worked on the motion.
 
Last edited:
hhjjy said:
Thank you . I have corrected it .
4. It works along half of the circle .
5. I am not sure , but I think the pitcher's arm is the only force on the ball. I am confuse that whether gravity has worked on the motion.
There is potential energy on the equation I posted above. If the applied force were zero, then the ball would still gain energy as it descends, right?
 
Yes , I can accept this concept .Potential energy is transferred to Kinetic energy.

Because of this equation ## \Delta Kinetic = Work +(-\Delta U) ## , we can know that both work and potential energy are increasing the kinetic energy . Right ?
PeroK said:
There is potential energy on the equation I posted above. If the applied force were zero, then the ball would still gain energy as it descends, right?
 
Last edited:
  • #10
hhjjy said:
Because of this equation ## \Delta Kinetic = Work +(-\Delta U) ## , we can know that both work and potential energy are increasing the kinetic energy. Right ?
Yes, both work by non-conservative forces and changes in potential energy can cause the kinetic energy of a body to change.

Remember that ##\Delta U_{\text{cons. force}} = -W_{\text{cons. force}}##, so your equation is equivalent to ##\Delta K = W_{\text{non-cons. forces}} + W_{\text{cons. forces}}##, which you should recognize as the work-energy theorem.

If you don't remember that, go back to your book and review how you transitioned from the work-energy theorem, where no distinction was made between conservative and non-conservative forces, to the concepts of potential energy and total mechanical energy.
 
  • #11
hhjjy said:
4. It works along half of the circle .
So do you think ## \frac \pi 2 ## in your solution is correct?
 
  • #12
I got it . It's ## W = F x =(30)( \frac{1}{2} 2 \pi r) ##. I forgot there is a 2 . Thank you .
 
  • Like
Likes   Reactions: hutchphd

Similar threads

  • · Replies 28 ·
Replies
28
Views
2K
Replies
55
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
11
Views
3K
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
2K
Replies
27
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K