Velocity at any point on a rotating sphere

In summary, the problem asks to find the current density at any point within a uniformly charged solid sphere that is spinning at a constant angular velocity. The solution is derived using the formula for current density, which is equal to the charge density multiplied by the velocity of the particles. The velocity of the particles is given by rωsin(θ) where r is the distance from the origin, ω is the angular velocity, and θ is the angle between the radial distance and the z-axis. This equation can be derived by considering the motion of particles on a rotating disk.
  • #1
Maxarthur
1
0

Homework Statement



"A uniformly charged solid sphere, of radius R and total charge Q, is centered at the origin and spinning at a constant angular velocity ω about the z axis. Find the current density [itex]\vec{J}[/itex] at any point (r,θ,[itex]\phi[/itex]) within the sphere."
Problem 5.6(b), p.223, from "Introduction to Electrodynamics" by Griffiths, 4th Edition.

Homework Equations



[itex]\vec{J}[/itex]=ρ[itex]\vec{v}[/itex].

The Attempt at a Solution


The solution is: [itex]\vec{J}[/itex]=ρrωsin(θ)[itex]\hat{\phi}[/itex], with [itex]\vec{v}[/itex]=rωsin(θ)[itex]\hat{\phi}[/itex], and ρ=[itex]\frac{Q}{(4/3)πR^{3}}[/itex].

Now, my real question: where did the expression for the velocity came from?
It makes sense that the velocity of any particle on a rotating sphere is given by [itex]\vec{v}[/itex]=rωsin(θ)[itex]\hat{\phi}[/itex], since v is proportional to r, points further away from the origin will be moving faster than those closer to it, ω is the radial velocity, and rω comes from the velocity of a particle in a rotating disk; rωsin(θ), θ being the angle between r and the z-axis, tells us that particles farther away from the z-axis will be moving faster than those closer to it: (assuming we're on the surface) since sin(0)=0 [itex]\Rightarrow[/itex] v=0 is the velocity of the point located at (0,0,R), which is not moving; and sin(π/2)=1 [itex]\Rightarrow[/itex] v is maximum, therefore all the points of any great circle are the fastest.
And [itex]\hat{\phi}[/itex] just tells us the direction in which the points of the sphere are moving.
Assuming my interpretation of the velocity equation is correct (correct me if wrong), I ask, where did the equation came from in the first place? I'm sure there has to be an analytical way of deriving such equation, and that it wasn't just a intuitive construction.
Thank you for your time, and let me tell you that I'm happy to belong to this community, this being my very first post (I apologize for any issues with the format of my comment); get used to see my username around here. :D

-Maxarthur

P.S: Every time that I hit the "Preview post" button, the "2. Homework Equations , ..." subsections get written on my message box. Why could this be? It's annoying having to erase them every time..
 
Last edited:
Physics news on Phys.org
  • #2
You can build a sphere by stacking an infinite number of co-centric circles with different radii on each other.Now if the sphere rotates around an axis passing through its poles,all those circles are doing so too with the same angular velocity but because the radii are different,the linear velocity of the points residing on the circles becomes different.[itex] r\sin{\theta} [/itex] is the formula for the radii of those circles which can be easily derived by drawing the shape of what I explained and checking the relation between the radius of the sphere and the radii of those circles.
 

1. What is meant by velocity at any point on a rotating sphere?

The velocity at any point on a rotating sphere refers to the speed and direction at which a specific point on the sphere's surface is moving as the sphere rotates around its axis. This velocity is constantly changing as the sphere rotates, and it is measured in units of distance per unit time, such as meters per second.

2. How is velocity at any point on a rotating sphere calculated?

The velocity at any point on a rotating sphere can be calculated using the formula v = ωr, where v is the linear velocity, ω is the angular velocity (measured in radians per second), and r is the distance from the point to the axis of rotation. This formula is based on the principle that all points on a rotating sphere have the same angular velocity, but their linear velocities vary based on their distance from the axis.

3. Does the velocity at any point on a rotating sphere change as the sphere rotates?

Yes, the velocity at any point on a rotating sphere is constantly changing as the sphere rotates. This is because the direction of the velocity vector changes as the sphere rotates, and the magnitude of the velocity depends on the distance from the axis of rotation. As a result, the velocity at any point on the sphere's surface is not constant.

4. How does the velocity at any point on a rotating sphere relate to its rotational speed?

The velocity at any point on a rotating sphere is directly proportional to its rotational speed, which is typically measured in revolutions per minute (RPM). This means that as the rotational speed increases, the velocity at any point on the sphere's surface also increases. Conversely, a decrease in rotational speed results in a decrease in velocity at any given point on the sphere.

5. What factors can affect the velocity at any point on a rotating sphere?

The velocity at any point on a rotating sphere can be affected by several factors, including the sphere's rotational speed, the distance of the point from the axis of rotation, and any external forces acting on the sphere (such as friction). Additionally, the shape and size of the sphere can also play a role in determining the velocity at any point, as well as any imperfections or irregularities in the sphere's surface.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
395
Replies
20
Views
891
  • Introductory Physics Homework Help
Replies
12
Views
199
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
183
  • Introductory Physics Homework Help
Replies
2
Views
255
  • Introductory Physics Homework Help
Replies
3
Views
218
  • Introductory Physics Homework Help
Replies
25
Views
272
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
126
Back
Top