Not sure about my rotations problem

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In summary, a rod of length L and mass M, initially balanced in a vertical position, tips over and rotates to the ground without slipping. The velocity of the center of mass of the stick just before it hits the ground is found to be v_f = L/2 * sqrt(3g/L). The velocity of the tip just before it hits is also found to be v_f = L * sqrt(3g/L). The center of rotation can vary depending on the frame of reference, but in this problem, it is clear that the bottom of the stick is the center of rotation. In general, the center of rotation is the point where an object rotates around, but in this case, the contact point with the ground is the
  • #1
toesockshoe
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Homework Statement


A rod of length L and mass M is balanced in a vertical position at rest. The rod tips over and rotates to the ground with its bottom attachment to the ground never slipping. Find the velocity of the center of mass of the stick just before it hits the ground and also find the velocity of the tip just before it hits.

Can someone check my work to see if I am correct?

Homework Equations


W=Delta E

The Attempt at a Solution


[itex] W=\Delta E [/itex]
[itex] 0 = \Delta RKE + \Delta GPE [/itex]
[itex] mg\frac{L}{2} = \frac{1}{2}Iw_f^2 [/itex]
[itex] mg\frac{L}{2} = \frac{1}{2} \frac{1}{3}mL^2w_f^2 [/itex]
[itex] w_f = \sqrt{\frac{3g}{L}} [/itex][/B]
 
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  • #2
You are doing fine so far. You now need to relate this angular frequency to the velocities the question asks for.
 
  • #3
Orodruin said:
You are doing fine so far. You now need to relate this angular frequency to the velocities the question asks for.
oh yeah, so just multiply it by r correct? so it owuld be: [itex] v_f = \frac{L}{2}\sqrt{\frac{3g}{l}} [/itex]
 
  • #4
This is for the center of mass, you should also do it for the tip (although the relation is trivial). I also suggest including the L/2 inside the square root (of course, it then becomes L^2/4) to simplify the expression slightly.
 
  • #5
Orodruin said:
This is for the center of mass, you should also do it for the tip (although the relation is trivial). I also suggest including the L/2 inside the square root (of course, it then becomes L^2/4) to simplify the expression slightly.

would you just multiply it by 2?
 
  • #6
Yes, the distance to the center of rotation is twice as large.
 
  • #7
Orodruin said:
Yes, the distance to the center of rotation is twice as large.
how would you prove this on a test, and what if they wanted the velocity of some point of b distnace away from the center? how would you find velocity then? multiply by b*2? Also, can you explain conceptually why its twice as fast at the tip? I just read it somewhere in my book so I said multiply by 2.
 
  • #8
Any point in a rotational motion has the velocity ##v = \omega r##, where r is the distance to the center of rotation. That the tip is twice as far away follows from the fact that the center of mass is in the middle of the rod.
 
  • #9
Orodruin said:
Any point in a rotational motion has the velocity ##v = \omega r##, where r is the distance to the center of rotation. That the tip is twice as far away follows from the fact that the center of mass is in the middle of the rod.
ok so why is the center of rotation at the bottom of the wheel? doesn't the wheel rotate around the middle of the wheel? a bicycle tire certainly looks like it spins around the center.
 
  • #10
toesockshoe said:
ok so why is the center of rotation at the bottom of the wheel? doesn't the wheel rotate around the middle of the wheel? a bicycle tire certainly looks like it spins around the center.

This depends on the frame you are using to look at the wheel. In the current problem, it is pretty clear what the center of rotation is.
 
  • #11
Orodruin said:
This depends on the frame you are using to look at the wheel. In the current problem, it is pretty clear what the center of rotation is.
ok yes, in this problem the center of rotation is the bottom of the stick. but imagine a bicycle wheel rolling... woudlnt it seem like the center of rotation in that case is the center of the bike?
 
  • #12
toesockshoe said:
ok yes, in this problem the center of rotation is the bottom of the stick. but imagine a bicycle wheel rolling... woudlnt it seem like the center of rotation in that case is the center of the bike?

If you use a system moving with the bike, yes. If you use a system fixed at the ground, no.
 
  • #13
Orodruin said:
If you use a system moving with the bike, yes. If you use a system fixed at the ground, no.
but it would be impossible to work with with a frame moving with the bike right? becuase wouldn't that be a noninertial frame?
 
  • #14
As long as the bike is not accelerating, the frame is inertial. And there is nothing preventing you from doing computations in non-inertial frames, it is just that you will have to take additional things into account.
 
  • #15
Orodruin said:
As long as the bike is not accelerating, the frame is inertial. And there is nothing preventing you from doing computations in non-inertial frames, it is just that you will have to take additional things into account.
ok assuming the bike is accelerating and I'm to dumb to use pseudo-forces, does the origin have to be the road?
 
  • #16
toesockshoe said:
ok assuming the bike is accelerating and I'm to dumb to use pseudo-forces, does the origin have to be some poinbt on the road?
 
  • #17
toesockshoe said:
ok assuming the bike is accelerating and I'm to dumb to use pseudo-forces, does the origin have to be the road?
If you are interested in the velocity of a point on the wheel relative to the road, it is the most convenient choice.
 
  • #18
Orodruin said:
If you are interested in the velocity of a point on the wheel relative to the road, it is the most convenient choice.
ok fine, but the center of rotation is still the middle of the wheel correct? we're just letting the ground be the orgin because its earier to solve problems?
 
  • #19
toesockshoe said:
ok fine, but the center of rotation is still the middle of the wheel correct? we're just letting the ground be the orgin because its earier to solve problems?

No. The (momentary) center of the rotation in the ground frame is the contact point (this is the point on the wheel with zero velocity).
 
  • #20
Orodruin said:
No. The (momentary) center of the rotation in the ground frame is the contact point (this is the point on the wheel with zero velocity).
ok that doesn't seem to make sense to me. can you explain what exactly a center of rotation is? i thought the center of rotation is the point where the object rotates around... a bicycle tire doesn't rotate around the ground or else it would also have to go under the ground right?
 
  • #21
toesockshoe said:
ok that doesn't seem to make sense to me. can you explain what exactly a center of rotation is? i thought the center of rotation is the point where the object rotates around... a bicycle tire doesn't rotate around the ground or else it would also have to go under the ground right?
The center of rotation is the point which for an object undergoing rotation would be (momentarily) at rest. This point does not necessarily lie within the object itself and may change from time to time.
 
  • #22
Orodruin said:
The center of rotation is the point which for an object undergoing rotation would be (momentarily) at rest. This point does not necessarily lie within the object itself and may change from time to time.
does it have to be at rest linearly as well as radially? in other words, if a bicycle is moving at constant speed, can the center of rotation of the tire be the center of the tire.
 
  • #23
toesockshoe said:
does it have to be at rest linearly as well as radially? in other words, if a bicycle is moving at constant speed, can the center of rotation of the tire be the center of the tire.

It has to be at rest, this means that the velocity is zero in all directions.
 

FAQ: Not sure about my rotations problem

1. What is a rotations problem?

A rotations problem involves determining the position, orientation, or movement of an object in a three-dimensional space. It can also refer to solving mathematical equations involving rotations or transformations.

2. How do I know if I have a rotations problem?

If you are dealing with an object that is moving or changing orientation in a three-dimensional space, or if you are trying to solve equations involving rotations or transformations, then you likely have a rotations problem.

3. What are the common approaches to solving rotations problems?

The most common approach is to use matrix operations, such as rotation matrices, to represent and manipulate rotations. Other methods include using quaternions, Euler angles, or geometric transformations.

4. How can I improve my understanding of rotations problems?

Practicing with different types of rotations problems and using visual aids, such as diagrams and animations, can help improve your understanding. Additionally, studying the underlying mathematical principles and concepts can also be beneficial.

5. Are there any real-world applications of rotations problems?

Rotations problems are used in various fields such as computer graphics, robotics, physics, and engineering. For example, in computer graphics, rotations are used to render 3D objects and create animations. In robotics, rotations are used to control the movement of robotic arms. In physics and engineering, rotations are used to study the motion of objects and analyze mechanical systems.

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