On the motion of a particle on the inner surface of a sphere

In summary, the conversation is about a physics problem involving a particle traveling from the bottom of a sphere to a peg with different velocities and reactions involved. The main equations and values for the velocities and reactions are given, with a final answer of h = 5a/4. However, there was an error in one of the equations, which was pointed out and corrected, resulting in a final answer of h = 7a/12.
  • #1
gnits
137
46
Homework Statement
Determination of the motion of a particle on the inner surface of a sphere
Relevant Equations
F = mv^2/r
Could anyone please help me out with the second part of this question:

246107

I've got the first part, u = √(5ga)

Here's my diagram for the second part:

246113

Distance traveled is from bottom of sphere to peg is 2πa/3 which means angle traveled is 2π/3.

So the particlee is going to travel 2π/3 radians and then hit the peg with velocity v1. It will rebound with velocity v2 after that will have a general position with velocity v3 and normal reaction R.

So, equating energy at start of motion at bottom of sphere with that at arrival at the peg we have:

½ m u^2 = ½ m v1^2 + mga

This rearranges to give:

v1 = √(u^2-2ga)

and as coefficient of restitution = ½ we then have:

v2 = ½ √(u^2-2ga)

So now, equate energy just after rebound with energy in general position to get:

½ m * ¼(u^2-2ga) = ½ m v3^2 + mga sin(ϒ)

This rearranges to give:

v3^2 = u^2/4 + ga/2 - 2ga sin(ϒ)

Now, centripetal force at general position is given by:

R + mg sin(ϒ) = m*v3^2/a

Substitute in for v3^2 from prior equation to get

R + mg sin(ϒ) = mu^2/4a + mg/2 - 2gma sin(ϒ)

Now we set R = 0 (this will be where the particle looses contact with the surface), then cancel through by m and substitute in that u = √(5ga) to get:

3g sin(ϒ) = 7g/4

This leads to sin(ϒ) = 7/12 = h / a and so

h = 7a/12 which would give an answer of 7a/12 + a = 19a/12

BUT the book asnwer is 5a/4

Thanks for any help,
Mitch.
 

Attachments

  • 9.JPG
    9.JPG
    24.4 KB · Views: 208
  • diagram.png
    diagram.png
    2.4 KB · Views: 204
  • diagram.png
    diagram.png
    2.4 KB · Views: 220
Physics news on Phys.org
  • #2
gnits said:
½ m * ¼(u^2-2ga) = ½ m v3^2 + mga sin(ϒ)

This rearranges to give:

v3^2 = u^2/4 + ga/2 - 2ga sin(ϒ)
Check the signs of the terms on the right side of the second equation above.
 
  • Like
Likes BvU
  • #3
Thanks very much TSny. I must have read and reread this error of sign every time I checked and rechecked. Indeed the answer then agrees with the books. Thanks very much.
 

Related to On the motion of a particle on the inner surface of a sphere

1. What is the equation for the motion of a particle on the inner surface of a sphere?

The equation for the motion of a particle on the inner surface of a sphere is given by r(t) = R sin(theta)cos(omega*t) i + R sin(theta)sin(omega*t) j + R cos(theta) k, where r(t) is the position of the particle, R is the radius of the sphere, theta is the angle of the particle from the center of the sphere, and omega is the angular velocity of the particle.

2. How does the angular velocity affect the motion of the particle on the inner surface of a sphere?

The angular velocity, represented by the variable omega, determines the speed at which the particle moves along the surface of the sphere. A higher angular velocity will result in faster motion, while a lower angular velocity will result in slower motion.

3. What is the significance of the radius of the sphere in the equation for the motion of the particle?

The radius of the sphere, represented by the variable R, plays a key role in determining the motion of the particle. It determines the size of the circular path that the particle follows and also affects the magnitude of the particle's velocity.

4. Can the particle's motion on the inner surface of a sphere be described as uniform circular motion?

Yes, the motion of the particle on the inner surface of a sphere can be described as uniform circular motion. This means that the particle moves at a constant speed along a circular path, with a constant radius and a constant angular velocity.

5. Are there any real-life applications of the motion of a particle on the inner surface of a sphere?

Yes, there are many real-life applications of the motion of a particle on the inner surface of a sphere. For example, it can be used to describe the motion of a pendulum or the motion of a planet around the sun. It can also be used in engineering and navigation systems to calculate the motion of objects on curved surfaces.

Similar threads

  • Introductory Physics Homework Help
2
Replies
40
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
949
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top