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

[physicsluv]

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 :(

 

[technician]

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 !

 

[BitterX]

you should keep the mass as 1.63kg
btw 1.63kg is 1630g

 

[technician]

Keep everything in kg, m and s

Do you know the answer?

 

[physicsluv]

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 :)

 

[technician]

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/

 

[physicsluv]

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.

 

[mtayab1994]

Is this the physics you're taking in college?

 

[technician]

that does not matter... the important thing is that you KNOW how to do it

 

[physicsluv]

Nope, normal high school physics.

 

[mtayab1994]

Nope, normal high school physics.

Yea that's what I thought as well because its really simple.

 

[physicsluv]

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

 

[technician]

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
Do you know what a slide rule is !

 

[mtayab1994]

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
Do you know what a slide rule is !

Is it a circular slide rule or a normal one?

 

[physicsluv]

Is it a circular slide rule or a normal one?

the acceleration and netforce is correct! and what is the slide rule?

 

[mtayab1994]

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.

 

[technician]

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.

 

[physicsluv]

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 :(

 

[technician]

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

 

[PeterO]

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

 

[technician]

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

 

[technician]

must replace my 'mechanical analogue calculator' with an electronic calculator... or get used to powers of 10

 

[PeterO]

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.

 

[BitterX]

try 1143.268637993

 

[PeterO]

try 1143.268637993

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