Circular Motion and Universal Law of Gravitation Help

In summary, the outside horse requires a greater net force to maintain its circular path compared to the inside horse on a merry-go-round, as the outside horse has a greater linear velocity due to its larger radius.
  • #1
rain_ex
6
0
1. In an amusement park ride, passengers stand inside an 8 m radius cylinder. Initially, the cylinder rotates with its axis oriented along the vertical. After the cylinder has acquired sufficient speed, it tilts into a vertical plane, that is, the axis tilts into the horizontal, as shown in the figure. Suppose that, once the axis has tilted into the horizontal, the ring rotates once every 4.5 s. If a rider's mass is 40 kg, with how much force does the ring push on her at the top of the ride?

knight_Figure_07_47.jpg


a. 390 N
b. 1000 N
c. 230 N *
d. 620 N

Can anyone check this for me? How I got this answer was: (I apologize for the lack of latex use)

- T = rev/time
- T = 2pi(radius)/4.5s
- T = 1.3956 radius/s
- N = (40kg)(1.9477radius/s)(8m) - (40kg)(9.8m/s^2) *used
- N = 231.262 N (closest to 230N, C)

2. Which requires a greater net force to maintain its circular path, a horse near the outside rail of a merry-go-round or a horse near the inside rail? *hint: the merry-go-round has the same rotational velocity at all points on the ride, but is the linear velocity the same for different radii?

a. the outside horse*
b. neither -- they both require the same net force
c. the inside horse

Can anyone check this for me? How I got this answer was:
The outside horse is obviously has a greater linear speed so that would mean the force has to be greater if both the radius and speed is greater, doesn't it?

3. From what height off the surface of Earth should an object be dropped to initially experience an acceleration of 0.5400 g's (5.292 m/s^2)? (Mass of Earth = 5.974 x 10^24 kg, Radius of Earth = 6.37 x 10^6 m )

a. 5.43 x 10^6 m
b. 1.69 x 10^6 m
c. 2.93 x 10^6 m
d. 3.56 x 10^6 m
e. 2.31 x 10^6 m

I don't really know where to start or how to solve this one. I tried using the universal law of gravitation but didn't know what to look for if the question refers to "height off the surface".

Help would be greatly appreciated. Thanks!
 
Physics news on Phys.org
  • #2
I agree with your answers for 1 and 2.
For #3, I suggest you start with
F ma = GMm/d²
Cancel the m's and you'll have a formula for the acceleration as it depends on the distance from the center of the Earth, d. You can solve it for d, then compute the distance that is above the Earth's surface.
 
  • #3
Okay I gave it another try and got D, 3.56x10^6 m, but I'm not certain if I followed you correctly.

Possible reasons for error:
Is "d" referring to the radius or is that what I substitute in?
Do I substitute "a" for 0.5400 or 5.292 m/s^2?

Thank you!
 
  • #4
"d" is the distance from the center of the Earth to the object.
The gravitational formula works on the distance between the centers of the two masses.

a = 5.292 or .5400*9.81
 
  • #5
I was wondering how did you know how to solve for the force the ring pushes on her?
 
  • #6
rain_ex said:
1.
3. From what height off the surface of Earth should an object be dropped to initially experience an acceleration of 0.5400 g's (5.292 m/s^2)? (Mass of Earth = 5.974 x 10^24 kg, Radius of Earth = 6.37 x 10^6 m )

a. 5.43 x 10^6 m
b. 1.69 x 10^6 m
c. 2.93 x 10^6 m
d. 3.56 x 10^6 m
e. 2.31 x 10^6 m

I don't really know where to start or how to solve this one. I tried using the universal law of gravitation but didn't know what to look for if the question refers to "height off the surface".

Help would be greatly appreciated. Thanks!

I trust you noted the acceleration was a little OVER one half what we experience on the surface.
Had it been exactly one half, the distance from the CENTRE of the Earth would have been 6.37 x 10^6 x 1.414 [root 2 times the Earth radius] which is 9.00 x 10^6
If it is was that far from the centre, it would be 2.63 x 10^6 off the surface.

Since the acceleration was a little OVER one half what we experience on the surface, it must be a little close - so I would expect (e) is the answer.

Calculate accurately and you will find that is correct. I have just made use of the multile choice format to save doing the accurate calculation.

had the options been 2.2, 2.3, 2.4, 2.5 , 2.6 I would have had to calculate carefully.

Peter

Interesting to note that if you did drop it from that height, you would have to have been in an orbiting spacecraft - far higher than the International Space Station - so having "dropped" it [ie released it] it would continue to orbit beside you - but with acceration towards the Earth of 0.54g, the same as your capsule.
Perhaps you have actually dropped it INSIDE the spacecraft - and you have probably seen video of the way objects just "float" in front of the astronaut then - you, the capsule and the object all accelerating towards the Earth at 0.54g., but not getting any closer!
 
  • #7
Boredgirl1 said:
I was wondering how did you know how to solve for the force the ring pushes on her?


The ring has radius 8m, and Period 4.5 seconds, so using the formula for Centripetal Force we have

F = 4π²Rm/T² we get 624 N.

At the top of the circle, that force is supplied by weight [down] plus Ring Push [down]

40 X 9.8 = 392 N, the ring applies the other 232 N which answer is closest to that? [after all I approximated g to be 9.8]

Peter
 
  • #8
rain_ex said:
2. Which requires a greater net force to maintain its circular path, a horse near the outside rail of a merry-go-round or a horse near the inside rail? *hint: the merry-go-round has the same rotational velocity at all points on the ride, but is the linear velocity the same for different radii?

a. the outside horse*
b. neither -- they both require the same net force
c. the inside horse

Can anyone check this for me? How I got this answer was:
The outside horse is obviously has a greater linear speed so that would mean the force has to be greater if both the radius and speed is greater, doesn't it?

If you use speed, you would be referring to the formula F = mv² / R

Now you have a larger v - [in the numerator], and a larger R [in the denominator]

Since the v term is squared while the R is not, the numerator could possibly win, and the force will be greater.

If you use the other formula for Centripetal Force

F = 4π²Rm/T²

You will see that all factors except R are equal for both horses - confirming that the horse "with the larger Radius" - the outside horse - needs the greater force.

Peter
 

1. What is circular motion?

Circular motion is the movement of an object in a circular path around a fixed point. It can be described as the continuous change in direction of an object's velocity, while its speed remains constant.

2. What is the Universal Law of Gravitation?

The Universal Law of Gravitation is a physical law that describes the force of gravity between two objects. It states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

3. How is circular motion related to the Universal Law of Gravitation?

In circular motion, the centripetal force that keeps an object in circular motion is provided by the gravitational force between the object and the center of rotation. This means that the Universal Law of Gravitation plays a crucial role in circular motion.

4. What is the difference between centripetal and centrifugal force?

Centripetal force is the inward force that keeps an object moving in a circular path, while centrifugal force is the outward force that appears to act on an object in circular motion. In reality, centrifugal force is not a real force, but rather an apparent force caused by the inertia of the object in motion.

5. How can the Universal Law of Gravitation be used to calculate the force of gravity between two objects?

To calculate the force of gravity between two objects, the masses of the objects and the distance between them must be known. Then, the force of gravity can be calculated using the formula F = G (m1m2/r^2), where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
661
  • Introductory Physics Homework Help
Replies
28
Views
1K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
222
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top