In summary, the conversation covers various statements and questions related to projectile motion, circular motion, and forces. Some statements are answered with certainty while others are left uncertain. The key concepts discussed include apparent weight, tension, radial and tangential acceleration, and the relationship between force, velocity, and acceleration. The conversation also emphasizes the importance of understanding these concepts for solving problems and performing well on tests.
  • #1
DanielleG
17
0

Homework Statement




  1. Answer True, False, or Cannot tell to each of the five statements below.
    1. A small projectile is launched horizontally 1 m above the surface of a smooth, airless planet, with sufficient speed for orbit. A bug riding in a small hole in the projectile has apparent weight equal to zero.
    2. A ball on a string moves in a vertical circle. When the ball is at its highest point, the tension in the string is less than the ball's weight.
    3. The figure shows the radial acceleration ar at four sequential points on the trajectory of a particle moving in a circle. The tangential acceleration at point 3 points up.
      circle2.gif
    4. All points on a rotating wheel have the same radial acceleration.
    5. A marble rolls around the inside of a cone. For the marble at the right side of the cone, the free body diagram is correct.
      marblecone.jpg
      fbd2.gif

Homework Equations



Not many equations for this, just theory

The Attempt at a Solution


1) True! Apparent weight = 0 N in free fall.
2) Unsure, but am leaning towards false.
3) Cannot tell! Depends on which way the particle is moving.
4) False, depends on the way the wheel is rotating.
5) False, Fc should be pointing to the left, not the right.

I have only one try left, and I'm just very unsure about b on the most part. Any guidance concerning #2? Thanks in advance!
 
Physics news on Phys.org
  • #2
For number two think what is actually happening when the ball is at the highest point. Draw a force diagram for the ball at the highest point and note what force is holding the ball down and what is holding it up (tension).
 
  • #3
For 3 and 4 you have the right answer but the reason you offer is wrong.
For 5, is it clear what the c in Fc stands for?
 
  • #4
RaulTheUCSCSlug said:
Draw a force diagram for the ball at the highest point and note what force is holding the ball down and what is holding it up (tension).

I've drawn a diagram and I know that at the top both the weight and tension are pointing down, but I am unsure of how to know which is smaller or larger.
 
  • #5
haruspex said:
For 3 and 4 you have the right answer but the reason you offer is wrong.

Hmmmm, okay, I'll give it some thought.

haruspex said:
For 5, is it clear what the c in Fc stands for?

I am basically positive it stands for centripetal force.
 
  • #6
DanielleG said:
I've drawn a diagram and I know that at the top both the weight and tension are pointing down,

Correct!

The sum of those two forces equals ...
 
  • #7
Mister T said:
The sum of those two forces equals ...

The centripetal force! So Fc = Ft + Fg, which can be rearranged to give Ft = Fc - Fg. Does that mean the tension in the string is greater than the ball's weight?
 
  • #8
DanielleG said:
The centripetal force! So Fc = Ft + Fg,

Correct!

which can be rearranged to give Ft = Fc - Fg.

I don't see how that rearrangement helps determine if Ft is less than Fg.

Does that mean the tension in the string is greater than the ball's weight?

Think about the ball's speed if Ft = 0. This is the minimum speed needed to maintain circular motion, a question sometimes asked on tests!
 
Last edited:
  • #9
Mister T said:
Think about the ball's speed if Ft = 0. This is the minimum speed needed to maintain circular motion, a question asked on test questions!

So the minimum speed needed to keep a ball in circular motion occurs when Ft = 0. Does that mean the weight is what's keeping the ball going in a circular motion? And if that's the case, then the tension in the string is definitely less than the ball's weight.
 
  • #10
DanielleG said:
So the minimum speed needed to keep a ball in circular motion occurs when Ft = 0. Does that mean the weight is what's keeping the ball going in a circular motion?

When the ball is at that position, yes.

And if that's the case, then the tension in the string is definitely less than the ball's weight.

And if the speed is larger than that minimum value Ft will be larger. Anything happen if Ft is larger yet and equal to Fg? Larger than that? What can you say about the ball's speed in each of these cases?

Thinking about these things now will definitely help you later when you're taking a test.
 
Last edited:
  • #11
Mister T said:
Anything happen if Ft is larger yet and equal to Fg? Larger than that? What can you say about the ball's speed in each of these cases?

Hmmm, would v get larger as Ft becomes larger?
 
  • #12
DanielleG said:
Hmmm, would v get larger as Ft becomes larger?

You got it!
 
  • #13
DanielleG said:
The problem statement
The figure shows the radial acceleration ar at four sequential points on the trajectory of a particle moving in a circle. The tangential acceleration at point 3 points up.

circle2.gif



The attempt at a solution

Cannot tell! Depends on which way the particle is moving.

Suppose the particle is moving in the clockwise direction. If the tangential component of the acceleration points upward the particle is speeding up. If downward the particle is slowing down.

If the particle is moving in the counter-clockwise direction we would have the opposite situation.

In general, if the directions of the velocity and the tangential component of the acceleration are the same, the particle is speeding up.

If the directions of the velocity and the tangential component of the acceleration are opposite, the particle is slowing down.
 
  • #14
DanielleG said:
The problem statement
All points on a rotating wheel have the same radial acceleration.

The attempt at a solution
False, depends on the way the wheel is rotating.

Points on the wheel that are different distances from the center will have different radial accelerations.

a
c = v2/r.
 
  • #15
Mister T said:
You got it!
Thank you!
 

Related to Centripetal Force Theory Questions

1. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and is necessary to maintain the object's circular motion.

2. How is centripetal force related to centripetal acceleration?

Centripetal force is directly proportional to centripetal acceleration. This means that as the force increases, the acceleration also increases. This relationship follows the equation F=ma, where F is the force, m is the mass of the object, and a is the acceleration.

3. Can centripetal force be calculated?

Yes, centripetal force can be calculated using the formula F=mv^2/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular path.

4. What is the role of centripetal force in the solar system?

In the solar system, centripetal force is responsible for keeping the planets in their orbits around the sun. The gravitational force between the sun and the planets acts as the centripetal force, keeping the planets in their circular paths around the sun.

5. Is centripetal force a real force?

Yes, centripetal force is a real force. It may not be a specific type of force, but rather a combination of other forces (such as gravity or tension) that act towards the center of the circular path. It is essential for circular motion and without it, an object would move in a straight line.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
8K
Replies
15
Views
2K
Back
Top