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

• hhjjy
In summary, the ball will gain kinetic energy as it descends due to the potential energy that was transferred to it from the equation above.
hhjjy
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.$$W = F x = 30N (\frac{1}{2}\pi r ) = 56.2 J$$
$$E_{system} = \Delta K + \Delta U = W$$
$$(K_{f}- K(i))+(U(f)-U(i)) = W$$
$$(\frac{1}{2} *m{V_{f}}^2 -\frac{1}{2}*m{V_{i}}^2)+(mgh_{f}-mgh{i}) = W)$$
$$V_{f} = \sqrt{V_{i}^2 + 2gh_{i} + \frac{2W}{m}}$$

1. Whether my solution is wrong.
2. What does the meaning of works done on a non-uniform circular motion?

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PhDeezNutz
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 .

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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 .

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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.

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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?

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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.

hhjjy said:
4. It works along half of the circle .
So do you think ## \frac \pi 2 ## in your solution is correct?

I got it . It's ## W = F x =(30)( \frac{1}{2} 2 \pi r) ##. I forgot there is a 2 . Thank you .

hutchphd

## 1. What is non-uniform circular motion?

Non-uniform circular motion is when an object moves in a circular path at varying speeds. This means that the object is not moving at a constant speed throughout the entire circular path.

## 2. How is work done calculated for non-uniform circular motion?

Work done for non-uniform circular motion can be calculated by taking the integral of the force applied to the object over the distance traveled. This takes into account the changing speed and direction of the object as it moves along the circular path.

## 3. Why is work done for non-uniform circular motion different from work done for uniform circular motion?

Work done for non-uniform circular motion is different from work done for uniform circular motion because in non-uniform circular motion, the speed and direction of the object is constantly changing, whereas in uniform circular motion, the object is moving at a constant speed throughout the entire circular path.

## 4. How does the radius of the circular path affect the work done for non-uniform circular motion?

The radius of the circular path does not affect the work done for non-uniform circular motion. Work done is only dependent on the force applied to the object and the distance traveled, not the shape or size of the path.

## 5. Can work done for non-uniform circular motion be negative?

Yes, work done for non-uniform circular motion can be negative. This occurs when the force applied to the object is in the opposite direction of the displacement, resulting in the object losing energy instead of gaining it.

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