Work Circular Motion Homework: Calculate Total Work Done on Ball

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Homework Help Overview

The discussion revolves around calculating the total work done on a ball of mass 0.765 kg, which is swung in a vertical circle. The problem includes multiple parts, focusing on the work done by tension and gravity during a complete circle and along a semicircle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of displacement being zero for parts a and b, while questioning the height change and energy involved in part c.
  • Some participants discuss the angle between the displacement and the force of tension, considering the nature of circular motion.
  • Questions arise about how to determine the angle and the relationship between displacement and tension.

Discussion Status

The discussion is ongoing, with participants examining the relationships between forces, displacement, and angles in the context of circular motion. Some guidance has been offered regarding the angle between tension and displacement, but no consensus has been reached on the overall approach to parts c and d.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use to arrive at conclusions.

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Homework Statement


A ball of mass 0.765 kg is tied to the end of a string of length 1.59 m and swung in a vertical circle.

a)During one complete circle, starting anywhere, calculate the total work done on the ball by the tension in the string.

b)During one complete circle, starting anywhere, calculate the total work done on the ball by gravity.

c)Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

d)Repeat part (b) for motion along the semicircle from the lowest to the highest point on the path.





The Attempt at a Solution



I know that for part a and b it will be 0, because the displacement is 0, according to W = F * s

However, I don't understand part c.

the displacement is 4.995.
 
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what is happening to the rock?? What is occurring to its height??

What kind of energy is being imparted to the stone as a result of this height change?

What is the height change?
 
what is happening to the rock?? it's moving up
What is occurring to its height?? increasing upwards

What kind of energy is being imparted to the stone as a result of this height change? gravity is pulling down, and tension is pulling it toward the center I believe

What is the height change? 4.995
 
Heat said:
what is happening to the rock?? it's moving up
What is occurring to its height?? increasing upwards

What kind of energy is being imparted to the stone as a result of this height change? gravity is pulling down, and tension is pulling it toward the center I believe

What is the height change? 4.995

For part c then
what is the angle between the displacement of the rock and the force of tension?? Consider very small displacements
 
the angle theta is 3.141509434
 
Heat said:
the angle theta is 3.141509434

why is that?
 
displacement is 4.995 right? (unless this is incorrect).

length of string is 1.59m

4.995m = 1.59(theta)
 
Heat said:
displacement is 4.995 right? (unless this is incorrect).

length of string is 1.59m

4.995m = 1.59(theta)

we'll talk about how you got the4.995 later.
How do you figure theta from there??

Think about it... the tension is pulling the rock inward. The rock is moving in a circle. If something is moving in a circle what is its instantaneous displacement?? Along what direction is the instantaneous displacement?? What is the angle between this instantaneous displacement and the force of tension??
 
if something is moving in a circle, then the displacement would be in radians. instantaneous displacement O_o

the direction of the instantaneous displacement is upwards.

The angle is pi/2 or 90 degrees.
 
  • #10
Heat said:
if something is moving in a circle, then the displacement would be in radians. instantaneous displacement O_o

the direction of the instantaneous displacement is upwards.

The angle is pi/2 or 90 degrees.

Good. SO the angle between the force of tension and the displacement is zero. Hence what is the work done by the tension?

W = F d \cos \theta = F d \cos \frac{\pi}{2} = ?
 
  • #11
0.

that would be part c.

now would part d be similar.
 

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