What is the Tension in a Vertical Circular Path with Given Parameters?

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Homework Help Overview

The discussion revolves around a physics problem involving an object in a vertical circular path, specifically focusing on calculating the tension in the string at the top of the circle. Participants are analyzing the forces acting on the object, including gravitational force and centripetal acceleration, while considering the parameters provided, such as mass, radius, and speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correct use of units and the implications of mass in calculations. There are attempts to clarify the calculations for centripetal acceleration and net force, with some questioning the accuracy of their results. Others express confusion about the methods used and seek further assistance on related parts of the problem.

Discussion Status

The conversation is ongoing, with various participants providing insights and corrections regarding unit conversions and calculation methods. Some have shared their results, while others are still grappling with the concepts and calculations involved. There is no explicit consensus on the final answers, but several productive lines of inquiry have emerged.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific requirements for accuracy and presentation of answers. There is mention of using different tools for calculations, such as slide rules and calculators, which may affect the discussion dynamics.

physicsluv
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Homework Statement


A 1.62 kg object is swung in a vertical circular
path on a string 1.8 m long.
The acceleration of gravity is 9.8 m/s^2.
If the speed at the top of the circle is 5 m/s,
what is the tension in the string when the
object is at the top of the circle?
Answer in units of N

Homework Equations


ac=mv^2/r
Fnet=ma

ac= Centripetal acceleration

The Attempt at a Solution

Ac=mv^2/2
ac=5^2/1.8
ac=13.89

(1.63kg=.00162g)
Fnet=ma
Ft+Fg=ma
Ft+(9.8*.00162)=13.89(.00162)
Ft=.006624

But that force of tension does not work :(
 
Last edited:
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You have the right idea but you have mixed up your units!
The mass is 1.62kg... no need to change it to anything else !
 
you should keep the mass as 1.63kg
btw 1.63kg is 1630g
 
Keep everything in kg, m and s

Do you know the answer?
 
Wow. I feel so stupid. That is like one of the basic parts of physics. I got 6.66488889 which was right. Thanks for your help :)
 
well done... I got 6.3 but I used g = 10 to make life easy !
A little tip... don't give so many figures in an answer... round it off to 3 figures/
 
haha! If only it was that easy, we are assigned problems online through UTexas, and so the answer has to be exact to get the question right.
 
Is this the physics you're taking in college?
 
that does not matter... the important thing is that you KNOW how to do it:smile:
 
  • #10
Nope, normal high school physics.
 
  • #11
physicsluv said:
Nope, normal high school physics.

Yea that's what I thought as well because its really simple.
 
  • #12
Can someone help me with part 3 of the problem?
For the acceleration I got 10.97670251m/s^2
and for the Net force I got 603.7186381 N

I did
fn-fg=fnet
and got
fn=1142.718638 and it was wrong
 

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  • #13
I got the acceleration to be 10.8... which is close enough to your answer... I do not have a calculator... I am using a slide rule:mad::wink:
Do you know what a slide rule is !
 
  • #14
technician said:
I got the acceleration to be 10.8... which is close enough to your answer... I do not have a calculator... I am using a slide rule:mad::wink:
Do you know what a slide rule is !

Is it a circular slide rule or a normal one?
 
  • #15
mtayab1994 said:
Is it a circular slide rule or a normal one?

the acceleration and netforce is correct! and what is the slide rule?
 
  • #16
physicsluv said:
the acceleration and netforce is correct! and what is the slide rule?

Uhm.. it's a mechanical analog calculator used for multiplication, division, and also logarithms and roots.
 
  • #17
Pluv
What a great question... what is a slide rule... I do have a calculator but the battery has gone flat... useless.
A slide rule is a great device for doing multiplications and divisions... be grateful that you will probably never need to resort to one.
 
  • #18
technician said:
Pluv
What a great question... what is a slide rule... I do have a calculator but the battery has gone flat... useless.
A slide rule is a great device for doing multiplications and divisions... be grateful that you will probably never need to resort to one.

Interesting! I am still at a loss as to how to find the force exerted by the pilot :(
 
  • #19
At the bottom of the loop the pilot is traveling at 210km/hr = 58.5m/s
Therefore the acceleration at the bottom of the loop = v^2/r = 11m/s^2
From this I got the force on the pilot to be 615N
 
  • #20
technician said:
At the bottom of the loop the pilot is traveling at 210km/hr = 58.5m/s
Therefore the acceleration at the bottom of the loop = v^2/r = 11m/s^2
From this I got the force on the pilot to be 615N

Assuming your acceleration figure of 11 is correct [I have no reason to doubt, but I have not calculated it myself] then the force you have calculated is the net force on the pilot [or close to it].

That force is achieved as the sum of two actual forces - gravity and the force from the seat.

Gravity pulls down with a force approaching 550N [if g=10 it is exactly 550 N]

The seat has to push strongly enough that the net force is your 615N up.

Draw some vectors to see what you get [if necessary]

EDIT: I just calculated it without rounding off on the way through and got 1143.3 N
 
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  • #21
PeterO
You are on the ball... I have made a slip in my calculations of x10
The acceleration at the bottom of the loop is 11m/s^2 which means that...
F-550N = 55 x 11 as the resultant force towards the centre of the circle
This gives F = (55 x 11 ) +550 = 610 + 550 = 1160N as the force on the pilot
... I have just seen your answer of 1143... will double check that I have not made 2 c ups
probably rounding off errors.
Thank you for your response
 
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  • #22
must replace my 'mechanical analogue calculator' with an electronic calculator... or get used to powers of 10
 
  • #23
technician said:
must replace my 'mechanical analogue calculator' with an electronic calculator... or get used to powers of 10

Yep! 55 x 11 = 605

and it should have been slightly less than 11 anyway.

Weight is 550N with g = 10, but you were told g = 9.81.
 
  • #24
try 1143.268637993
 
  • #25
BitterX said:
try 1143.268637993

Nice numbers, but needs a bit of work to be the answer to this question.
 

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