2 Questions: Centripetal Force related

In summary, centripetal force is the force that keeps an object moving in a circular path and is directly related to circular motion. The formula for calculating centripetal force is F = m * v^2 / r, and as the radius of the circular path increases, the centripetal force decreases. Centripetal force can be greater than the weight of an object due to its dependence on mass, velocity, and radius rather than weight.
  • #1
Thorlax402
1
0
Question 1:
In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 7.00 rad/s, as in the figure displayed below. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? (Hint: Recall that the magnitude of the maximum force of static friction is equal to µn, where n is the normal force - in this case, the force causing the centripetal acceleration.)


3. The Attempt at a Solution : 14.9847 (Obviously Wrong, way too large for a coefficient.
Basically, my problem on this one comes down to solving for centripetal force without knowing the mass. I can easily get centripetal acceleration, but don't know where to go from there with the data given. If someone could explain how to do this that would be fantastic.


Question 2:
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 34.0 m/s. With what maximum speed can it go around a curve having a radius of 71.0 m?

My answers (both wrong): 71.8310 m/s, 16.0933 m/s
Quite frankly, I don't know what I am doing wrong on this one. For the first of my two answers, I thought I was being given angular velocity which is not the case, but the second one not only used tangential velocity like I was supposed to, but the answer makes sense and is still not right. If someone could help me on this one too it would be greatly appreciated.


Thanks in advance,
~Thorlax
 
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  • #2
Answer 1:The minimum coefficient of friction needed to keep a rider from slipping is µ = 0.48.To solve this problem, we need to calculate the normal force n that is causing the centripetal acceleration. This is given by the equation n = mv^2/r, where m is the mass of the rider, v is the angular velocity, and r is the radius of the cylinder. Plugging in the given values, we get n = m*(7.00 rad/s)^2/3.00 m = 14.98 m/N. Now we can calculate the minimum coefficient of friction needed to keep the rider from slipping. The magnitude of the maximum force of static friction is equal to µn, so µ = 0.48.Answer 2:The maximum speed with which the truck can go around a curve having a radius of 71.0 m is 16.09 m/s.We can calculate the maximum speed of the truck by using the equation v = √(g*r), where g is the gravitational constant (9.8 m/s^2) and r is the radius of the curve. Plugging in the given values, we get v = √(9.8 m/s^2 * 71.0 m) = 16.09 m/s.
 
  • #3

I would approach these questions by first identifying the relevant equations and variables involved. For question 1, we can use the equation for centripetal force, Fc = mv^2/r, where m is the mass of the rider, v is the tangential speed, and r is the radius of the cylinder. We also know that the maximum force of static friction is given by µn, where µ is the coefficient of friction and n is the normal force. In this case, the normal force is equal to the rider's weight, mg, since the riders are suspended against the wall. Therefore, we can set up the following equation:

Fc = µmg

Substituting in the values given in the problem, we get:

(mv^2/r) = µmg

We can rearrange this equation to solve for µ:

µ = (mv^2)/(mgr)

However, we still do not know the value of m. To solve for this, we can use the fact that the tangential speed, v, is equal to the angular speed, ω, multiplied by the radius, r. Therefore, we can rewrite the equation as:

µ = (mω^2r)/(mgr)

The mass, m, cancels out and we are left with:

µ = ω^2r/g

Now, we can plug in the values given in the problem to solve for µ:

µ = (7.00 rad/s)^2(3.00 m)/(9.8 m/s^2)

µ = 14.29

This is the minimum coefficient of friction needed to keep the rider from slipping. It is important to note that this is a very high value and may not be achievable in real-life situations.

For question 2, we can use the same equation for centripetal force, Fc = mv^2/r. However, in this case, we are given the radius and we need to solve for the maximum speed, v. We can set up the following equation:

Fc = mv^2/r

Since the mass, m, is the same for both curves, we can set the two equations equal to each other and solve for v:

(mv^2)/r = (mvmax^2)/rmax

Solving for vmax, we get:

vmax = (v√(r/rmax)

Plugging in the values given in the problem, we get
 

Related to 2 Questions: Centripetal Force related

1. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, always pointing towards the center of the circle. It is responsible for keeping the object moving along the circular path instead of flying off in a straight line.

2. How is centripetal force related to circular motion?

Centripetal force is directly related to circular motion as it is the force that keeps an object moving along a circular path. It is necessary for an object to maintain circular motion, otherwise, the object would move in a straight line due to inertia.

3. What is the formula for calculating centripetal force?

The formula for calculating centripetal force is F = m * v^2 / r, where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

4. How does centripetal force change with the change in the radius of the circular path?

As the radius of the circular path increases, the centripetal force decreases. This is because the centripetal force is inversely proportional to the radius. In other words, the force required to keep an object moving in a larger circle is less than the force required to keep it moving in a smaller circle.

5. Can centripetal force be greater than the weight of an object?

Yes, centripetal force can be greater than the weight of an object. This is because centripetal force is not dependent on the weight of the object, but rather on its mass, velocity, and the radius of the circular path. In certain situations, the centripetal force needed to maintain circular motion may be greater than the weight of the object.

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