Using v= omega cross r to find instantaneous velocity

Click For Summary
To find the instantaneous velocity of an object rotating at 4 rad/s about a specified axis, the cross product formula \(\vec{v} = \vec{\omega} \times \vec{r}\) is utilized. The angular velocity vector was calculated as \(\vec{\omega} = (1.486 \hat{i} - 2.971 \hat{j} + 0.557 \hat{k})\) rad/s, and the position vector \(\vec{r}\) was determined as \((-1, 2, -3)\). The resulting velocity vector was computed, but concerns were raised about the low value of the k component and the potential for rounding errors. It was suggested to retain the prefactor in its original form to avoid complications and rounding issues. The method used was confirmed to be correct, allowing for further calculations with the original surd values.
FaraDazed
Messages
347
Reaction score
2

Homework Statement


An object is rotating at 4 rad/s about an axis in direction of (2 \hat{i} - 4 \hat{j} + 3\hat{k} ) which then passes through a point (1,2,0)m .

Calculate the instantaneous velocity at the point (2,0,3)m (Hint: use \vec{v} = \vec{\omega} \times \vec{r})

Homework Equations


<br /> \hat{A} = \frac{\vec{A}}{|\vec{A}|}<br />

The Attempt at a Solution


First off, i would like to say I am extremely new (1-2 weeks new) to cross products, and this is the first time we have been given physical (rather than purely mathematical) problems to implement it.

What I immediately thought of was the equation I put in the relevant equations section, how a unit vector is calculated and since all I need/want from the vector (2 \hat{i} - 4 \hat{j} + 3\hat{k} ) is the direction I turned that into a unit vector, and then to get the angular velocity vector times that by 4. So...
<br /> \sqrt{2^2+4^2+3^2} = \sqrt{29} \\<br /> \therefore \vec{\omega} = (\frac{4 \cdot 2}{\sqrt{29}} \hat{i} - \frac{4 \cdot 4}{\sqrt{29}} \hat{j} \frac{4 \cdot 3}{\sqrt{29}} \hat{k}) rad/s = (\frac{8}{\sqrt{29}} \hat{i} - \frac{16}{\sqrt{29}} \hat{j} + \frac{12}{\sqrt{29}} \hat{k}) rad/s
And then converted to decimal to makes things easier
<br /> \vec{\omega} = (1.486 \hat{i} - 2.971 \hat{j} + 0.557 \hat{k}) rad/s<br />

And then to get \vec{r} I did (1,2,0)-(2,0,3)=(-1,2,-3)

And then did the cross product, I am not sure how to do matrices in latex but I put i,j,k on top row (of 3 by 3 matrx) and then on second row put the omega vector and then on third put (-1,2,-3).

Then I got the determent and thus the velocity vector to be...
<br /> \vec{v} = [(-2.971 \cdot -3)-(0.557 \cdot 2)] \hat{i} + [(1.486 \cdot -3)-(0.557 \cdot -1)] \hat{k} + [(1.486 \cdot 2)-(-2.971 \cdot -1)] \hat{k} \\<br /> \vec{v} = 7.8 \hat{i} -3.9\hat{k}+0.001\hat{k}<br />

I don't know if my method is correct at all, I am suspicious of the low value for the k component for a start off. Oh and as the question asks for "the instantaneous velocity" I do not know whether they mean just the velocity vector or its magnitude; I assume if they meant the magnitude then they would have just said "speed" instead, though.
 
Physics news on Phys.org
FaraDazed said:
And then converted to decimal to makes things easier
This makes things a lot more complicated. It is easier to keep the prefactor ##\frac{4}{\sqrt{29}}## as it is (don't even put it to the components). It stays a common prefactor of everything, so you can continue to work with the nice integers (like 2,-4,3 for your rotation axis) and care about the ugly prefactor later. This also does not lead to the rounding error you see in your result.

They probably mean velocity, but you can give both. You have the velocity, the speed is easy to calculate.
 
mfb said:
This makes things a lot more complicated. It is easier to keep the prefactor ##\frac{4}{\sqrt{29}}## as it is (don't even put it to the components). It stays a common prefactor of everything, so you can continue to work with the nice integers (like 2,-4,3 for your rotation axis) and care about the ugly prefactor later. This also does not lead to the rounding error you see in your result.
ok, fair enough, thanks. But is my method correct and what I did after converting to decimal? Just so if I know its correct I can just go back an keep with the surds but follow the same steps.
 
Yes the approach is correct.
 
mfb said:
Yes the approach is correct.

Ok, thanks for your help :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
2
Views
792
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
Replies
1
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
11
Views
2K