Circular Motion?

Question

In physics we're doing a unit over circular motion and I'm so confused. I don't understnad it at all. Could you please explain these to me? Thanks!

In a carnival the "Spinner of Death" is a ride where all the participants stand in metal reinforced ring with diameter of 10m. THe ride then accelerates the ring to a certain angular velocity such that the participants are pressed against the wall. THe floor of the ride is then dropped and the participants remain pressed agains the wall. Coefficeint of friction is .5. a) What angualr velocity must the ride achieve in order to keep the participants from falling off the ride? b) If the floor falls after 2 minutes, what must angular acceleration be in order for this to happen?

A 100,000 N truck takes a curve w/ a radius of 200m. The truckis traveling 75 km/h and coefficient of friction at that spot is .09. Will the truck remain on the road or slide off? Why or why not.

Thank you for any explanations.

Answer 1

Carnival Problem a)

centripetal acceleration: a = rw^2

Since F = ma,

centripetal force: F = m rw^2

The centripetal force will equal how hard the ride pushes on the riders - very much like a normal force is how hard a surface pushes on an object. Since friction is mu (coefficient of friction) x how hard two surfaces are pushed together, you can say

friction = mu x m x rw^2

Now this friction is the only thing keeping the rider from sliding down, so it has to be equal to the weight (mg) of the rider, so set them equal to each other.

mu x m x rw^2 = mg (mass cancels out)

Rearranging for w (angular speed) gives

w = sqroot(g / (r x mu))
so
w = sqroot(9.8 m/s^2/(5 m x .5))
w = 1.98 rad/s

b) w i = inital angular velocity, w f = final, alpha = angular accelaration

wi = 0 , wf = 1.98 rad/s

wf = wi + alpha x t

w f / t = alpha

alpha = 1.98 rad/s / 120 s
alpha = .0165 rad/s^s

Truck problem:

First convert 75 km/hr to m/s

75 km/h x 1000 m / 1 km x 1 h / 3600 s = 20.8 m/s

Next a = v^2 /r
a=20.8^2/200 = 2.17 m/s^2

To know how much force the truck needs to make the turn, we need to multiply the mass of the truck x the acceleration (F = ma)

The mass of the truck is 100,000 N / 9.8 m/s^s = 10204 kg

So F = 10204 kg x 2.17 m/s^2 = 22140 N

The force of friction available to make the truck turn is the coefficient of friction x weight of the truck (which is the same as the normal force acting on the truck).

Friction = .09 x 100,000 N = 9000 N

So there's much less friction (9000 N) than is needed (22140 N) to make the turn, so it will slide off.

Answer 2

Its called centripetal force. I'll use the classic example of a rock on a string:

Take a rock and tie it to a string. Hold onto the string and drop the rock. Gravity pulls down on the rock but the rock can only drop as far as there is string left. Now swing the string in a circle at your side, as you move the string in a circular motion, the rock appears to go in a circular pattern. As the rock moves it wants to fly away but the string keeps it at bay. Gravity is kept in check by the force that is pushing the rock to the end of the string. If you cut the string as the rock is moving, the rock will shoot away from you. When you stop twirling the string, the rock stops moving and falls down.

Its all about the effects of gravity and force.

In the Spinner of death, the faster you spin the more force you create. At a certain rate you can drop the floor and the force of the spinning will keep you pinned against the wall. Here, the wall is the rope and you are the rock.

In the case of the truck, gravity is pushing down on the truck as he takes the turn. If the speed is not over a certain point, the truck maintains contact with the road and makes the turn.

As its been a very long time since I did the math on this stuff, I can't provide an exact answer. Hope fully, this will help what the problem really is about.

Answer 3

If you look at a FBD of a rider, the forces on the rider in the vertical are:

gravity = m*g downward
static friction = N*.5 upward

N is related to the angular velocity w as
N=m*w^2*r
where r is the radius

you plug this into the equations governing vertical motion:
.5*m*w^2*r-m*g=0

this is the minimum w to balance the forces. Any greater angular velocity and the person is just pressed more into the wall and is more safe against slipage.

plug and crunch
w=sqrt(2*g/r)

let's use 10 m/s^2 for g
=sqrt(2*10/10)
=sqrt(2) rad/s

the angular acceleration average is simpley
w/(2*t)
in this case
sqrt(2)/(2*2*60) rad/s^2

The second is similar
the force holding the truck on the curve is friction
N*.9
100,000*.9
90,000 newtons

the force pushing against friction is
m*w^2*r

in this case we're given the linear speed v
v=w/r
so
m*v^2/r is the force
(100,000*(75/3.6)^2)/200
217,014 newtons

this is greater, truck slides off the road.

j