Circular Motion and calculating time

[ForgottenMemo]

Homework Statement

Homework 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

The Attempt at a Solution

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

Is T(Time) = 360 radians?

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

 

[Hootenanny]

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

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

 

[ForgottenMemo]

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

What do you mean.

1) Transpose the equation again

or

2) there's another formula that i have missed


If so, any hint on where to look.

 

[Hootenanny]

What do you mean.

1) Transpose the equation again

or

2) there's another formula that i have missed


If so, any hint on where to look.

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

 

[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

 

[Hootenanny]

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

I'm not checking your arithmetic, but your method looks good

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

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?

 

[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

 

[Hootenanny]

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

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?

 

[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)

 

[Hootenanny]

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)

Looks okay to me

 

[ForgottenMemo]

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

 

[Hootenanny]

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

Not a problem