Circular Motion and calculating time

1. Jun 24, 2008

ForgottenMemo

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

2. Relevant equations

Velocity =2PiR/T

Time=2PiR/V

Force Centripetal = MV^2/R

Angular Displacement/Time interval = Average Angular Velocity

Angular Velocity x Radius Of Circle Path = Linear Velocity

Average Angular Acceleration = Change In Angular Acceleration/ Time Interval

Linear Acceleration = Radius of Circular Path x Angular acceleration

Force= Mass x Acceleration

3. The attempt at a solution

How do i calculate time? Once i know that i can use this formula: Velocity =2PiR/T

Im not honestly sure where to start, ive been trying for 4 days now, constantly seeking infomation while at work.

2. Jun 24, 2008

Hootenanny

Staff Emeritus
Wouldn't it perhaps be easier to use a different formula, one in which you know all the required information?

3. Jun 24, 2008

ForgottenMemo

What do you mean.

1) Transpose the equation again

or

2) Theres another formula that i have missed

If so, any hint on where to look.

4. Jun 24, 2008

Hootenanny

Staff Emeritus
The formula is on your list . You know the centripetal force, the mass and the radius. You want to know the velocity.

5. Jun 24, 2008

ForgottenMemo

I so hope this is correct.

Fc= MV^2/R -> Fc x R/M=V^2

Velocity -> (1100x140)/1000 = 154^2 -> $$\sqrt{154}$$ = 12.4 (m/s ?)

Acceleration -> V^2/R -> 12.4^2/140 -> 1.09 m/s^2

Force Gravity -> M x g -> 1000x9.81 = 9810g

Last edited: Jun 24, 2008
6. Jun 24, 2008

Hootenanny

Staff Emeritus
I'm guessing by the amount of g acting on the car, they mean how many times the acceleration due to gravity is the acceleration of the car, rather than the actual force. Do you follow?

7. Jun 24, 2008

ForgottenMemo

Would i be correct in saying that:

Acceleration x Gravity = The amount of g acting on the vehicle?

1.09 m/s^2 x 9.81 = 10.69

8. Jun 24, 2008

Hootenanny

Staff Emeritus
Not quite. If you accelerate at 1g means you are accelerating at a rate that is equal to the acceleration due to gravity (i.e. 9.81 m.s-2). Similarly, an acceleration of 2g is equivalent to an acceleration of twice that of gravity (i.e. 19.62 m.s-2). That number of g's 'pulled' can be calculated thus:

$$n = \frac{a}{g}$$

Do you follow?

Last edited: Jun 24, 2008
9. Jun 24, 2008

ForgottenMemo

Ok that would mean that.

Acceleration / Gravity = The amount of g acting on the vehicle

1.09 m/s^2 / 9.81 = 0.1 (g)

10. Jun 24, 2008

Hootenanny

Staff Emeritus
Looks okay to me

11. Jun 24, 2008

ForgottenMemo

Hootenanny : It was a pleasure. Thanks for the help.

12. Jun 24, 2008

Hootenanny

Staff Emeritus
Not a problem