Centripetal force problems for dynamics

[pinkyboo]

Centripetal force problems for dynamics :)

Homework Statement

I have two questions!

1. You are driving at a speed of 60.0km/h around a turn with a radius of 75m. The coefficient of friction between the tires and the road is 0.85.
a. After traveling 20° around the turn, you hit a patch of ice and your coefficient of friction is reduced to 0.15. After 5 seconds, where are you with respect to the start of the turn?
b. Repeat question a. with a banked curve of 35°.

2. A 2.00kg ball is dropped, it sticks to a rope, and makes a circular motion (see below). What is the net force on the banana at the bottom of the swing. (Assume the velocity is constant once it hits the rope... or, for more accuracy, use the conservation of energy formulas)

Homework Equations

Fnet = ma_c

a_c = mv²/r

a_c = 4∏²r/T²

The Attempt at a Solution

(first question)

v = 16.6666667m/s, r = 75m, μ = 0.85

For this one, I'm not even really sure where to start. :(

I wanted to do the Fnet = m x 4∏²r/T² equation, but since I don't have the period, I wasn't really sure where to go at all. Furthermore, part b just boggles me

I also wanted to find F_N or just do something concerning the normal or friction forces since I do have the coefficient of friction, but again, without the mass, I'm at a loss.

(second question)

I attached the diagram of the question.

m = 2.00 kg

Grrrr... with this one, I tried to make an FBD, but all attempts were futile! I tried to make Fgx in the direction of the motion, but I'm at a loss for this question as well.


Sorry my attempts at any solutions seem so juvenile, but I really am not sure what to do. Any help would be greatly appreciated! :rofl:

 

[Mentz114]

First calculate the angular velocity ω = v/r. Then the centripetal force is F_c = m ω² r. The frictional force is F_f = μ m g. So we have eqilibrium if m ω² r < μ m g. g is the gravitational acceleration downwards and r and v are the radius and velocity. Notice that m cancels out.

Can you take it from here ?

 

[pinkyboo]

Thank you! I get now how you canceled the masses, but as for 'ω'... I just searched it, and though I do understand it, it's not something we learned in my school. :( would I be able to use F^c = m x 4∏²r/T²?

And after this step, I'm at 451.828058 m/s with the velocity...

and do you have any idea for how I would go about solving parts a and b? I'm really bad at this. :( sorry for all the questions!

 

[Mentz114]

Angular velocity is the rate an angle(in radians) is changing. It is 1/(2 π T).

Problem a)

1. Find the centripetal force.
2. Find the frictional force for μ=0.85
3. Same as 2 for μ=0.15
4. get 1-3. This is the force now making the car slide

Think about it. Draw the car at the point it starts to slide and resolve the forces.

 

[Nickie]

Hello there! It seems like you have some interesting questions about centripetal force and dynamics. Let's start with the first question.

1. The first step in solving this problem is to find the centripetal force acting on the car as it goes around the turn. We can do this by using the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the turn.

a. So, in this case, Fc = (m)(v^2)/r = (m)(16.6666...^2)/(75) = 3.704 m/s^2. This is the force that must be provided by the friction between the tires and the road in order for the car to stay on the turn.

Now, when the car hits the patch of ice, the coefficient of friction decreases to 0.15. This means that the frictional force decreases as well. We can find the new frictional force by using the equation Ff = μN, where Ff is the frictional force, μ is the coefficient of friction, and N is the normal force.

Since we know that Fc = Ff, we can set these two equations equal to each other and solve for N. This will give us the normal force that the car needs in order to maintain the same centripetal force as before.

Fc = Ff
(m)(v^2)/r = μN
N = (m)(v^2)/(rμ)

Now, we can plug in the values to find N: N = (m)(16.6666...^2)/(75)(0.15) = 111.11 N.

This means that the car needs a normal force of 111.11 N in order to maintain the same centripetal force as before. But since the coefficient of friction is now lower, the normal force will also be lower. This means that the car will start to slide outwards from the turn.

To find the position of the car after 5 seconds, we need to use the equation x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

In this case, we know that x