Bar suspended by rope kinetic energy

In summary, the total kinetic energy of the bar suspended by a rope at one of its ends (also known as a bar pendulum) is given by the translational energy calculated at the center of mass plus the rotational energy around the center of mass. This can be represented by the equation T = (1/2)*m*(v^2_cm) + (1/2)*I*(w^2), where m is the mass of the bar, v_cm is the speed of the center of mass, I is the moment of inertia around the center of mass, and w is the angular velocity. In the specific case of a bar oscillating in a 2D plane, the kinetic energy can be calculated by T = (
  • #1
rmfw
53
0
If I needed to calculate the kinetic energy of a bar suspended by a rope at one of its ends, in other words, a bar pendulum, do I need to calculate the kinetic energy of the center of mass plus the kinetic energy relatively to the center of mass or do I need to calculate the kinetic energy of the point located on the beginning of the bar (bar-rope junction) plus the kinetic energy relatively to that point? I'm thinking the first way is the right way, thanks.

Edit: Only the bar has mass, so when I speak of kinetic energy and center of mass I'm referring to the bar only.
 

Attachments

  • picture.jpg
    picture.jpg
    6.2 KB · Views: 385
Last edited:
Physics news on Phys.org
  • #2
Also sorry for that doublepost, didn't meant to.
 
  • #3
rmfw said:
If I needed to calculate the kinetic energy of a bar suspended by a rope at one of its ends

Zero if the bar is not moving, but if that were the problem I don't suppose you'd be asking :smile:

And kidding aside, you'll have to tell us more about the problem (which, depending on little details like whether the rope is "ideal" or not, may be seriously hairy) and especially the motion that the bar is undergoing.

At any given moment the kinetic energy of the bar is the sum of the translational kinetic energy ##\frac{mv^2}{2}## calculated at the centter of mass of the bar and the rotational kinetic energy ##I\omega^2##. However, these will change in possibly non-trivial ways as the rope jerks the bar around.
 
  • Like
Likes 1 person
  • #4
I have added a picture to the original post.

Ok so [itex] T = \frac{ m v_{centerofmass}^2}{2} + \frac{I w^2}{2} [/itex]

So now I still don't know if I should use [itex] I = \frac {m L^2}{12} [/itex] (rotation about the center of mass) or [itex] I = \frac {m L^2}{3} [/itex] (rotation at the beginning of the bar)
 
Last edited:
  • #5
Ok I already know the answer is [itex] I = \frac{m L ^2}{12}[/itex] . But there's something else bothering me, if there were no rope, only the bar fixed at its beginning (origin) and oscillating in a 2D plane, then the kinetic energy would be only in rotational motion calculated using [itex] I = \frac{m L ^2}{3}[/itex] , what differs from one case to the other? In this second example the center of mass is also moving and there's also rotation relatively to the center of mass. This is really confusing me.
 
  • #6
You're doing a pretty good job at answering your own questions. Just keep going. Kidding aside, Kidding aside, The total kinetic energy is given by the tanslational energy (using the speed of the center of mass) plus the rotational energy around the center of mass.
 
  • Like
Likes 1 person
  • #7
Ok I understood that, but my question still remains, (see pic attached in this post) in this case, I know the answer for kinetic energy is [itex] T = \frac{1}{2} \dot{\theta}^2 \frac{m L^2}{3} = \dot{\theta}^2 \frac{m L^2}{6}[/itex]

But, according to this statement "The total kinetic energy is given by the tanslational energy (using the speed of the center of mass) plus the rotational energy around the center of mass. "

Why isn't it [itex] T= \frac{m V_{cm}^2}{2} + \frac{1}{2} \dot{\theta}^2 \frac{m L^2}{12} = \frac{m (\frac{L}{2} \dot{\theta})^2}{2} + \frac{1}{2} \dot{\theta}^2 \frac{m L^2}{12}= 4 \dot{\theta}^2 \frac{m L^2}{24} = \dot{\theta}^2 \frac{m L^2}{6} [/itex]

Dear lord, what kind of sorcery is this? I should have checked the results before even asking it. Thanks anyway, I get it now
 

Attachments

  • picture 2.png
    picture 2.png
    967 bytes · Views: 387

What is a bar suspended by rope kinetic energy?

A bar suspended by rope kinetic energy refers to the potential energy stored in a bar that is suspended by a rope and is in motion due to the force of gravity.

How is the kinetic energy of a bar suspended by rope calculated?

The kinetic energy of a bar suspended by rope can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the bar and v is the velocity at which it is moving.

What factors affect the kinetic energy of a bar suspended by rope?

The kinetic energy of a bar suspended by rope is affected by the mass of the bar, the velocity at which it is moving, and the height at which it is suspended.

What happens to the kinetic energy of a bar suspended by rope as it moves?

The kinetic energy of a bar suspended by rope decreases as it moves due to the conversion of kinetic energy into potential energy. As the bar moves downwards, its potential energy increases and its kinetic energy decreases.

How is the kinetic energy of a bar suspended by rope related to its potential energy?

The kinetic energy and potential energy of a bar suspended by rope are directly related, as they both involve the movement of the bar. As the kinetic energy decreases, the potential energy increases and vice versa.

Similar threads

Replies
3
Views
852
Replies
3
Views
933
Replies
13
Views
1K
Replies
5
Views
787
Replies
5
Views
874
Replies
4
Views
934
  • Mechanics
Replies
3
Views
1K
Replies
30
Views
2K
Replies
9
Views
2K
Back
Top