What is the correct moment of inertia for a spinning disk?

[ac7597]

Homework Statement

The Jupiter 6 is a giant spaceship in the shape of a big, flat disk: it has radius R=5000 m, height h=200 m, and mass M=8.00E+08 kg. As it flies through space, it rotates around its center with a period of P=200 seconds to generate artificial gravity. The rotation looks counter-clockwise as seen from above the ship.
One day, one of the outboard thruster rockets malfuctions. It is located at the point marked with a big black dot on the diagram, on the outer rim of the ship, an angular distance θ=30 degrees away from the x-axis. The thruster exerts a force with components 71î +0ĵ −158k̂ mega-Newtons for a duration of t=2 seconds before the technicians can disable it.

What is the torque exerted around the center of the ship by the thruster in the î direction, in the ĵ direction, and in the k̂ direction ?

What is the new rotation period of the ship after the thruster stops firing?

Relevant Equations

torque= |force||radius|sin(theta)

Since torque= |force||radius|sin(theta) :
î = |71E6||5000m|sin(30)= 177.5E9 N*m
ĵ = |0||5000m|sin(30) = 0
k̂ = |-158E6||5000m|sin(30)= 395 N*m

 

[collinsmark]

Homework Statement: The Jupiter 6 is a giant spaceship in the shape of a big, flat disk: it has radius R=5000 m, height h=200 m, and mass M=8.00E+08 kg. As it flies through space, it rotates around its center with a period of P=200 seconds to generate artificial gravity. The rotation looks counter-clockwise as seen from above the ship.
One day, one of the outboard thruster rockets malfuctions. It is located at the point marked with a big black dot on the diagram, on the outer rim of the ship, an angular distance θ=30 degrees away from the x-axis. The thruster exerts a force with components 71î +0ĵ −158k̂ mega-Newtons for a duration of t=2 seconds before the technicians can disable it.

What is the torque exerted around the center of the ship by the thruster in the î direction, in the ĵ direction, and in the k̂ direction ?

What is the new rotation period of the ship after the thruster stops firing?
Homework Equations: torque= |force||radius|sin(theta)

Since torque= |force||radius|sin(theta) :
î = |71E6||5000m|sin(30)= 177.5E9 N*m
ĵ = |0||5000m|sin(30) = 0
k̂ = |-158E6||5000m|sin(30)= 395 N*m

You're mixing and matching your methods of determining cross products.

It's true that

$\vec \tau = \vec r \times \vec F = |r||F| \sin \theta,$

but that's only useful if you know what $\theta$ is (the angle between $\vec r$ and $\vec F$ [It's not 30 deg here]), have already calculated the magnitudes of $r$ and $F$, and even then, you'll still need to calculate a new angle for the one corresponding to $\vec \tau$. But you haven't done all that yet (I don't recommend using this approach, btw. See below).

[Edit: Actually, I take that back. The force and radius vectors do line up nicely on the x-z-plane. So using the above approach is fine. You'll still need to calculate $\theta$ and the magnitudes of $r$ and $F$ though. That said, I still think that the approach below is probably still easier anyway.]

The above approach is convenient if both the radius and the force line up nicely on a convenient plane like the x-y-plane. But that's not the case here.

Another way to find the cross product (which I do recommend in this case is):

$$\vec \tau = \vec r \times \vec F = \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$$

Your fist order of business then is to calculate the components of $\vec r$.

 

[ac7597]

rx=5000m
ry=5000m
rz=200m

 

[jbriggs444]

rx=5000m
ry=5000m
rz=200m

That point is not even on the ship.

Recall that torque is the cross product of a vector-valued point of application ($\vec{r}$) and a vector-valued force ($\vec{F}$). We are looking for the coordinates of the point of application, $\vec{r}$.

 

[collinsmark]

rx=5000m
ry=5000m
rz=200m

As @jbriggs444 mentions, something's not quite right there.

You're looking for the coordinates where the force is applied [relative to the ship's center of mass]. Since the ship's center of mass is at the origin: the coordinates of "the point marked with a big black dot on the diagram."

 

[collinsmark]

It might be worth pointing out that the coordinate system defined in the diagram has the y-axis in the "up - down" direction (not the z-axis), as seen from the viewer's perspective. (This threw me at first too in my earlier response, before my edit.)

So the plane of the ship lies mostly in the x-z-plane (not the x-y-plane).

Also, while not specifically indicated, I think it's safe to assume that the thruster rocket lies along the center of the height of the ship (still on the edge of the ship, but in the center of the 200 m height). In other words, I think it's safe to assume that $r_y = 0$. So that just leaves $r_x$ and $r_z$ to calculate.

 

[ac7597]

Rx = -5000 m
Ry = 0
Rz = -200 m

 

[collinsmark]

Rx = -5000 m
Ry = 0
Rz = -200 m

No, I think you're just guessing now.

$r_x \hat \imath$ and $r_z \hat k$ are at right angles to each other. It's possible to put them together in such a way that they form a right triangle. The hypotenuse of that right triangle is 5000 m. How does one find the components of such a triangle? (Hint: this is where the 30 deg comes in.)

 

[ac7597]

5000sin(30) = 2500 rx
5000cos(30) = 4330 rz

 

[collinsmark]

5000sin(30) = 2500 rx
5000cos(30) = 4330 rz

That's closer to the right idea, but still not quite. (Not to mention It looks like you forgot a couple of "=" signs.)

Look carefully at the diagram. Notice that $\theta$ is defined with respect to the x-axis.

So, does that make $r_x$ the "opposite" leg of the triangle or the "adjacent" leg?

----------------------------------
For future reference, recall:

$\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$

$\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

$\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}}$

 

[ac7597]

rx = 4330
rz = 2500 ?

 

[collinsmark]

rx = 4330
rz = 2500 ?

Yes! Correct.

But I think it's important you understand why.

$\vec r = 4330 \hat \imath + 0 \hat \jmath + 2500 \hat k \ [\mathrm{m}]$

Take that equation and look at the diagram with the disk on it. Follow out each component along the x-axis, y-axis, and z-axis on the diagram. Do you see how that shows the coordinates of the "big black dot"?

If not, try to concentrate on that until you can see it. It may be important for future problems.

This particular problem made it a little easier, because all the values are non-negative here. You might want to ask yourself, "what if $\theta$ was going in the other direction (so there's a component along the negative z direction instead of the positive z). How would that have affected the sign of the components?

Anyway, these are just ideas to help familiarize yourself with 3D vectors.
----------------------

So, now all that's left is to plug them into

$$\vec \tau = \vec r \times \vec F = \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$$

and evaluate.

 

[ac7597]

î = 0 N*m
ĵ = 861.6E9 N*m
k̂ = 0 N*m
Thanks!
I confused on how to find the rotational period. Do I need to find the angular acceleration: torque= (moment inertia) * (angular acceleration). Then use angular acceleration to find the angular velocity : ω1=0+ (angular acceleration)(2seconds). Linear velocity= ω1*radius.
Finally, period=(2*pi*radius)/Linear velocity

 

[collinsmark]

î = 0 N*m
ĵ = 861.6E9 N*m
k̂ = 0 N*m
Thanks!

Good work!

I confused on how to find the rotational period. Do I need to find the angular acceleration: torque= (moment inertia) * (angular acceleration). Then use angular acceleration to find the angular velocity : ω1=0+ (angular acceleration)(2seconds). Linear velocity= ω1*radius.
Finally, period=(2*pi*radius)/Linear velocity

Yes, that approach will work fine. But realize, there's no need to go back to linear velocity at this point. Recall that the period $T$ is just the inverse of the simple frequency $f$.

$f = \frac{1}{T}$

And the angular velocity (a.k.a., angular frequency) $\omega$ is

$\omega = 2 \pi f$

A small bit of substitution will easily relate $T$ and $\omega.$

------------------

There's another way to conceptually think about solving the rest of this, although the actual calculations are almost identical to what you described above.

In linear mechanics, the impulse (force times time) equals change in linear momentum.

There is an angular version that you can use here, for this problem:

Torque times time equals change in angular momentum.

But whether you use the angular impulse method here, or the angular acceleration method you described above, you should get the same answer.

 

[collinsmark]

Do I need to find the angular acceleration: torque= (moment inertia) * (angular acceleration). Then use angular acceleration to find the angular velocity : ω1=0+ (angular acceleration)(2seconds).

Oh, and just to be clear, the torque applied for those 2 seconds will apply a change to the angular velocity. Don't forget it was spinning before rocket misfired (initially, the angular velocity was non-zero). So you'll need to add this new change in angular velocity to the initial angular velocity (rather than add it to 0).

 

[ac7597]

Iy= (1/12)(mass)(3*radius^2 + height^2)
Iy= (1/12)(8*10^8)(3*5000^2 + 200^2)= 5*10^15
torque= 861.6E9 N*m = 5*10^15 *(angular acceleration)
angular acceleration= 172.3 * 10^-6
ω1= 0 + 172.3 * 10^-6 (2second) = 344.6E-6
ω1= 344.6E-6 =2π *f
f=54.8E-6
period= (54.8E-6 )^-1= 18.2 * 10^3 seconds
For some reason this is wrong

 

[collinsmark]

Iy= (1/12)(mass)(3*radius^2 + height^2)
Iy= (1/12)(8*10^8)(3*5000^2 + 200^2)= 5*10^15
torque= 861.6E9 N*m = 5*10^15 *(angular acceleration)
angular acceleration= 172.3 * 10^-6
ω1= 0 + 172.3 * 10^-6 (2second) = 344.6E-6
ω1= 344.6E-6 =2π *f
f=54.8E-6
period= (54.8E-6 )^-1= 18.2 * 10^3 seconds
For some reason this is wrong

Look up the moment of inertia for a solid disk/cillinder spinning around its primary axis. (I think you used one for thin rod)

 

[ac7597]

w0= 2*pi/ (200s) =31.4E-3
correction: ω1= 31.4E-3 + 172.3 * 10^-6 (2second) = 31.76E-3
31.76E-3/(2*pi) = f = 5.05E-3
period = (5.05E-3)^-1 = 197.8 seconds

 

[ac7597]

Iy= (1/4) (mass) (radius) ?
Isn't there a height for the disk?

 

[collinsmark]

Hold, on. Could you step back a bit and walk me through it. Something is really wrong with the Moment of Inertia formula that you used.

(Hint: the 200 m height of the ship is irrelevant. If you're using that in your calculations somewhere then something isn't right.)

 

[collinsmark]

Iy= (1/4) (mass) (radius) ?
Isn't there a height for the disk?

From what I gather from the problem statement, you are supposed to treat the ship as a solid disk.

Here is a useful link for moments of inertia:
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

 

[collinsmark]

Okay, I see what you did.

Pay close attention to which axis the disk is spinning. The disk is spinning around its central axis -- the one that has cylindrical symmetry.

You were using an axis as through it's tumbling top over bottom. That's not the axis you want here.

 

[ac7597]

Iy= (1/4)(mass)(radius^2)
Iy= (1/4)(8*10^8)(5000^2)= 5*10^15
torque= 861.6E9 N*m = 5*10^15 *(angular acceleration)
angular acceleration= 172.3 * 10^-6
w0= 2*pi/ (200s) =31.4E-3
ω1= 31.4E-3 + 172.3 * 10^-6 (2second) = 31.76E-3
31.76E-3/(2*pi) = f = 5.05E-3
period = (5.05E-3)^-1 = 197.8 seconds
I got the same answer even if I used either the Iy= (1/4)(mass)(radius^2) or Iy= (1/12)(mass)(3*radius^2 + height^2)

 

[collinsmark]

Iy= (1/4)(mass)(radius^2)
Iy= (1/4)(8*10^8)(5000^2)= 5*10^15
torque= 861.6E9 N*m = 5*10^15 *(angular acceleration)
angular acceleration= 172.3 * 10^-6
w0= 2*pi/ (200s) =31.4E-3
ω1= 31.4E-3 + 172.3 * 10^-6 (2second) = 31.76E-3
31.76E-3/(2*pi) = f = 5.05E-3
period = (5.05E-3)^-1 = 197.8 seconds
I got the same answer even if I used either the Iy= (1/4)(mass)(radius^2) or Iy= (1/12)(mass)(3*radius^2 + height^2)

The moment of inertia formula is not (1/4)(mass)(radius^2) for a spinning disk, spinning about its axis of [cylindrical] symmetry.

 

[ac7597]

Iy= (1/4)(8*10^8)(5000^2)= 5*10^15
Iy= (1/12)(8*10^8)(3*5000^2 + 200^2)= 5*10^15
I confirmed that 197.8s is correct

 

[collinsmark]

Iy= (1/4)(8*10^8)(5000^2)= 5*10^15
Iy= (1/12)(8*10^8)(3*5000^2 + 200^2)= 5*10^15

The correct moment of inertia of a disk depends on which axis it is spinning.

If a solid disk spins like a normal wheel on a car, a record on a turntable, or pretty much any disk that you would imagine spinning like a normal disk, its moment of inertia is

$I = \frac{1}{2}mr^2$

On other other hand, if you grab a disk by its edges, with one hand on one side and the other hand on the other side of the disk, and you use your fingers to spin it end over end, then its moment of inertia is

$I = \frac{1}{12}m \left(3 r ^2 + h^2 \right)$

But that's not the moment of inertia that you want in this problem.

For this problem you want to use the $I = \frac{1}{2}mr^2$ formula. The spaceship disk is initially spinning normally (like most disks do). And when the torque is applied it doesn't change that. The torque is applied on the same axis as the initial rotation.

I confirmed that 197.8s is correct

If you used either

$I = \frac{1}{4} mr^2$

or

$I = \frac{1}{12} m \left(3 r^2 + h^2 \right)$

then you got the wrong answer. An online automated program might have "accepted" the answer to be within reasonable limits, but it's still technically wrong (I imagine the program would have accepted any answer near 200 seconds).