You are Here: Home >< Physics

Announcements Posted on
Last day to win £100 of Amazon vouchers - don't miss out! Take our quick survey to enter 24-10-2016
1. Hello all,

Attached is a screenshot of the question I am struggling with conceptually, and was hoping someone could clarify the physics:

So this is what I understand so far:
to solve for travel time involves simple harmonic motion about the COM of the space station *superposed* with centripetal force contribution due to eth stations Earth orbit. Otherwise SHM alone physically doesn't permit the bead to move further away from the COM than it's original displacement.

The SHM part is relatively straightforwards, it's the centrifugal part that would cause the bead to be forced outwards, like whirling a loaded shopping bag around. But strangely centrifugal 'force' hasn't been explicitly covered in my course so far and various internet sources disagree as to it's precise definition. So input from one of the wise members on here would be appreciated.

Edit RE centrifugal: I realise that it's a manifestation of N1L and is NOT a force, but I struggle to convince myself why it occurs at all. For example a centrifugal switch in an AC motor.
2. Wow. That's a bit of a mind bender.

First, I am not convinced the gravitational coupling of the bead and the space station is necessary. If you were to model that, the space station would also oscillate in anti-phase and it would be a mess. It wouldn't even be SMH since the restoring force due to gravitational attraction is ~1/r^2, i.e. not proportional to the displacement. So my suggestion is to disregard that.

What really matters is the initial displacement of the bead from the COM of the space station. You can reduce the problem to one of two beads, one (space station) at exactly the right orbital height to remain in a circular orbit (216km)-i.e. centripetal force equals gravitational force-and another one (the bead) is just a little bit too high to remain in circular orbit (216.001km)-i.e. the centripetal force (force required to ensure circular motion) exceeds the gravitational force.

In reality, that means the bead will appear to 'accelerate' outwards. In reality it is still accelerating inwards but not fast enough to stay in a circular orbit. The amount of 'acceleration', which I cautiously call centrifugal acceleration is the difference between the centripetal force required and the actual gravitational force. So let's figure each of these out.

centripetal force required: F_req = m*r*w^2

We have no idea what the mass is. But have faith.

gravitational force F_grav = mg

Centrifugal force is then the difference and centrifugal acceleration is a = F/m (Newton's 2nd):

F_cfg = mrw^2 - mg
a_cfg = rw^2 - g

The problem is we don't know w, the angular speed of the orbit. But we do have similar equations for the space station: applying the same equations to that (and using ' to distinguish it from the bead):

F_req' = m'*r'*w^2
F_grav' = m'g

Note that g is the same (approximation!) and w is the same because the bead is stuck on the wire. Since the space station actually is in a circular orbit, these two forces are equal (once again, F_req is not really a force, it's the force required to satisfy the condition of circular motion).

m'*r'*w^2 = m'g

r'*w^2 = g

so w^2 = g/r'.

Now we can sub that into the expression for the centrifugal acceleration:

a_cfg = rw^2 - g

a_cfg = r(g/r' ) - g = g(r/r' - 1)

Which is quite delightful when you look at it. When the bead is slightly further away from the Earth and r > r', so r/r' > 1, and a_cfg is positive. When the bead is slightly closer to the Earth, r < r', so r/r' < 1 and a_cfg is negative. Exactly what we'd expect in orbital mechanics. This looks like a reasonable answer under the approximation that there is negligible change in gravitational field strength with height (i.e. (r'-r)/r' << 1 so so (g(r)-g(r' )/g(r' ) << 1). I leave it to you to sub in and, presumably, take a stab at the depth of the Earth (since r and r' are measured to the Earth COM!).
3. That is very helpful, thanks so much.
(Original post by mik1a)
In reality, that means the bead will appear to 'accelerate' outwards. In reality it is still accelerating inwards but not fast enough to stay in a circular orbit. The amount of 'acceleration', which I cautiously call centrifugal acceleration is the difference between the centripetal force required and the actual gravitational force. So let's figure each of these out.
This is the crux of the question for me, and what I could not fathom.
Therefore RE the underlined sentence, is this the 'definition' of a centrifugal acceleration when studying *any* rotating frame?
Therefore on say a merry-go-round does friction with the ride floor replace eh gravity in this question?
4. Yep, that's right.

In a nonrotating frame of reference, where your axes are fixed in space, F=ma.

In a rotating frame, where your axis rotate in space, F - F_cfg = ma, where F_cfg is an apparent force that contributes to the mass's acceleration in the rotating frame only.

This term appears because a, the acceleration of the object, is now being measured with reference to a rotating set of coordinates rather than a fixed set. When you transform from fixed to rotating coordinates, a few interesting new terms are spewed out of the equations of motion. Like F_cfg, which is not a real force. The centrifugal force is just the force that would ordinarily explain the weird motions you see because your reference frame is rotating.

In addition to the centrifugal force, the Coriolis force is also spewed out (applies to moving objects), as is the lesser-spotted Euler force (which only appears when the rotating frame's spin rate changes). You can derive them all if you start with a rotating reference frame, (x',y') = R(x,y), where R is your time-dependent rotation matrix, and try to derive the velocity and acceleration of an object with position vector u(x,y) in the (x',y') frame. They all just spew out.
5. (Original post by mik1a)
Yep, that's right.

In a nonrotating frame of reference, where your axes are fixed in space, F=ma.

In a rotating frame, where your axis rotate in space, F - F_cfg = ma, where F_cfg is an apparent force that contributes to the mass's acceleration in the rotating frame only.

This term appears because a, the acceleration of the object, is now being measured with reference to a rotating set of coordinates rather than a fixed set. When you transform from fixed to rotating coordinates, a few interesting new terms are spewed out of the equations of motion. Like F_cfg, which is not a real force. The centrifugal force is just the force that would ordinarily explain the weird motions you see because your reference frame is rotating.
Great that helps.

However I am still a bit lost. Here is my understating of centrifugal (I will refer to it as a pseudo force from here on in to distinguish it from actual forces) 'force':

An observer in the rotating frame introduces the centrifugal 'force' () in order to to be able to use Newton's laws in a non-inertial frame (consistent with N1L).
, on a mass m, is defined from the motion of the rotating observer (called O' wrt stationary inertial observer O), on a circle of radius r:
for O'

...BUT how can an inertial observer O, e.g. just floating in space watching the satellite and bead orbiting Earth, see this centrifugal pseudo force causing the bead to accelerate outwards. This contradicts being in an *inertail* frame?
Similarly how does this explain why a centrifugal governor on a steam engine gets wider/arms accelerate outwards as increases from 0, as the engine speeds up?

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: September 24, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

### Q&A with Paralympian Jack Rutter

Poll
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.