# Circular Motion- Vertical

1. Dec 15, 2011

### physicsluv

1. The problem statement, all variables and given/known data
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

2. Relevant equations
ac=mv^2/r
Fnet=ma

ac= Centripetal acceleration
3. 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: Dec 15, 2011
2. Dec 15, 2011

### 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 !!

3. Dec 15, 2011

### BitterX

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

4. Dec 15, 2011

### technician

Keep everything in kg, m and s

Do you know the answer?

5. Dec 15, 2011

### 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 :)

6. Dec 15, 2011

### technician

well done.... I got 6.3 but I used g = 10 to make life easy !!!!
A little tip..... dont give so many figures in an answer... round it off to 3 figures/

7. Dec 15, 2011

### 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.

8. Dec 15, 2011

### mtayab1994

Is this the physics you're taking in college?

9. Dec 15, 2011

### technician

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

10. Dec 15, 2011

### physicsluv

Nope, normal high school physics.

11. Dec 15, 2011

### mtayab1994

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

12. Dec 15, 2011

### 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

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13. Dec 15, 2011

### 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:grumpy:
Do you know what a slide rule is !!!!!!

14. Dec 15, 2011

### mtayab1994

Is it a circular slide rule or a normal one?

15. Dec 15, 2011

### physicsluv

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

16. Dec 15, 2011

### mtayab1994

Uhm.. it's a mechanical analog calculator used for multiplication, division, and also logarithms and roots.

17. Dec 15, 2011

### 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.

18. Dec 15, 2011

### physicsluv

Interesting! I am still at a loss as to how to find the force exerted by the pilot :(

19. Dec 15, 2011

### technician

At the bottom of the loop the pilot is travelling 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. Dec 15, 2011

### PeterO

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

Last edited: Dec 15, 2011
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