Hello it's a me! Today we continue with some more examples in Physics.
The last two posts where about examples in:
Object equilibrium and 2nd law applications
Today's topic will be circular dynamics and vertical circle motion examples.
The theory for that can be found here and I highly suggest you to check it out!
So, without further do, let's get started!
Example 1 (flat curve):
Suppose we have a flat/horizontal curve on a highway with an radius (r) of 200m. A car takes this corner with a velocity of 20 m/s. Which is the least possible (static) friction factor μs so that the car doesn't slide out?
In constant velocity circular motions we know that the acceleration is a = u^2/r.
That way from the second law we have:
ΣF = F = ma = mu^2/r =>
F = m*20^2/200 =>
F = 2m N
The maximum static friction is also equal to:
F = μs*Ν = μmg = 10μm N
By setting those equations equal to each other we get:
2m = 10μsm =>
μs = 0.2
Example 2 (banked curve):
Suppose that we now have a banked curve on a highway with an radius (r) of 350m that will be constructed with an angle (θ). A car needs to be able to take this corner with a velocity of 25 m/s and the friction is theoretically supposed as zero (non-existent). Which is the least angle (θ) so that this can happen?
In my theory post we already found the equations but here a small explanation
The vertical ground force has an angle θ to the gravity and so we find the components:
Nsinθ and Ncosθ (one for each axis)
Using the 1st and 2nd law for each axis respectively we get:
N * sinθ = mu^2 / r (2nd law)
N * cosθ = mg (1st law)
By dividing these equations we get:
tanθ = u^2/gr =>
θ = arctan(u^2/gr) => θ = arctan(25^2/3500) ~= arctan(0.1785) =>
θ ~= 10.12 degrees
Example 3(vertical circle):
The radius (r) of a theme park wheel is 9m and it does a circle in 12 seconds.
Calculate the phenomenal weight of an 80kg passenger at the highest and lowest point of the wheel.
Because the period is T = 12 seconds we can calculate the velocity from:
u = 2πr / T => u ~= 4.71 m/s
The centripetal force Fc stays the same and so:
Fc = mu^2/r => Fc != 197 N
Because the gravity also stays the same and is W = mg = 800 N we know that there is a upward force at the top that is declining the gravity so that:
Fc = W - Ft = mu^2 / r => Ft = W - Fc =>
Ft = 800 - 197 = 603 N (upwards)
This force is acting like a vertical ground force and so the phenomenal weight the top is 603N!
In the same way at the bottom there is a upward force that is also declining gravity so that:
Fc = Fb - W = mu^2 / r => Fb = W + Fc =>
Fb = 197 + 800 = 997N (upwards)
So, the phenomenal weight at the bottom is 997N!
Another interesting topic is how we generate artificial gravity by rotating a space station in space!
You can read about that here.
And this is actually it and I hope that you learned something!
Next time we will get into more advanced Newton law examples and after that we are finished with everything that I wanted to cover about Classical mechanics (for now)! This means that we will get into a new physics branch!