Calculating Centripetal Force: Tension in a Whirling Stone on a String | HW Help

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Homework Help Overview

The discussion revolves around calculating the tension in a string caused by a stone being whirled horizontally. The problem involves concepts of centripetal force, mass, radius, and velocity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of equations related to centripetal force and express uncertainty about which equation to apply. There are attempts to clarify the importance of using consistent units and writing equations symbolically before substituting values.

Discussion Status

Some participants have provided guidance on ensuring unit consistency and have encouraged posting equations for verification. There is an ongoing exploration of different equations and approaches to the problem, but no consensus has been reached on the correct method.

Contextual Notes

Participants note the need to convert units to the mks system and express confusion about the appropriate equations provided by their teacher. There is also mention of a worksheet context, indicating that this is part of a structured assignment.

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Homework Statement


1. Centripetal Force: If a 40.0 gram stone is whirled horizontally on the end of a 0.60 m string at a speed of 2.2 m/s. What is the tension in the string?



3. I did F = 40/0.60 X 2.2. Is that right?
 
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PsychoPowell said:

Homework Statement


1. Centripetal Force: If a 40.0 gram stone is whirled horizontally on the end of a 0.60 m string at a speed of 2.2 m/s. What is the tension in the string?



3. I did F = 40/0.60 X 2.2. Is that right?

Not quite right, but on the right track. Write out the equation you are using symbollically first. Be careful not to drop squared terms. Also be sure to keep your units consistent. Do not mix meters with grams, for example. Keep/convert everything into mks (meters-kilograms-seconds) units.

EDIT -- Welcome to the PF, BTW!
 
berkeman said:
Not quite right, but on the right track. Write out the equation you are using symbollically first. Be careful not to drop squared terms. Also be sure to keep your units consistent. Do not mix meters with grams, for example. Keep/convert everything into mks (meters-kilograms-seconds) units.

EDIT -- Welcome to the PF, BTW!
Teacher gave us a bunch of equations, so I'm not sure which is the right equation.

Also, thanks for the Welcome. :smile:
 
PsychoPowell said:
Teacher gave us a bunch of equations, so I'm not sure which is the right equation.

Also, thanks for the Welcome. :smile:

There are two main equations that you will use for centripetal force (the subject you are studying now). The wikipedia page may clear it up for you a bit. You are given things like linear velocity, mass and radius in your problem statement, I believe...

http://en.wikipedia.org/wiki/Centripital_force

.
 
berkeman said:
There are two main equations that you will use for centripetal force (the subject you are studying now). The wikipedia page may clear it up for you a bit. You are given things like linear velocity, mass and radius in your problem statement, I believe...

http://en.wikipedia.org/wiki/Centripital_force

.
This is one:

75f3b2a3b8aabe3f09d3637dd0cd5f19.png


Not sure how I should do it though.
 
PsychoPowell said:
This is one:

75f3b2a3b8aabe3f09d3637dd0cd5f19.png


Not sure how I should do it though.

Not sure what that is. How about trying the first "formula" on the wikipedia page that I linked for you? Be sure to use all mks units...
 
Yeah, the first formula should solve the problem. I got 322.66.
 
PsychoPowell said:
Yeah, the first formula should solve the problem. I got 322.66.

Be sure to include units in your equations and results. It helps to prevent math errors and misconceptions.

Also, if you post your equations and work, we can usually check them here. If you already submitted the answer and it was correct, you don't have to do that for this question.
 
berkeman said:
Be sure to include units in your equations and results. It helps to prevent math errors and misconceptions.

Also, if you post your equations and work, we can usually check them here. If you already submitted the answer and it was correct, you don't have to do that for this question.
It's a work sheet. I did 40 X (2.2)^2/.60
 
  • #10
PsychoPowell said:
It's a work sheet. I did 40 X (2.2)^2/.60

Please post the full equation with units. It's honestly meaningless without units. You will learn this in your coming years, grasshopper...
 
  • #11
berkeman said:
Please post the full equation with units. It's honestly meaningless without units. You will learn this in your coming years, grasshopper...
Haha sorry.

40 grams X (2.2 m/s)^2/0.60 m = 322.66
 
  • #12
PsychoPowell said:
Haha sorry.

40 grams X (2.2 m/s)^2/0.60 m = 322.66

grams? Ack.
 

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