Kinetic Energy of Rigid Object Swinging in Vertical Plane

In summary, the conversation discusses the total kinetic energy of a rigid object swinging in a vertical plane. The use of a stretchable cord and the changing center of the bar are considered when calculating the kinetic energy. The parallel axis theorem is also mentioned in relation to the moment of inertia. The conversation also touches on the inclusion of the term for x_c and when it is necessary to consider the translation kinetic energy for rotation. The conversation concludes with a mention of using a mass instead and its potential differences.
  • #1
KFC
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I am quite confusing about the total kinetic energy of an rigid object swing in a vertical plane. Imagine a bar hanging to the ceiling with a massless but stretchable cord. The bar has mass M. Because the cord is stretchable (not very hard but like a spring with very small spring constant), the center of the bar [tex](x_c, y_c)[/tex] changed.

Well if the cord is NOT stretchable, since the bar is swinging, the kinetic energy is [tex]K = \frac{I}{2}\omega^2[/tex]. In this case, the center of the bar is also changing from place to place, but we don't account [tex]\frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)[/tex] into the total kinetic energy. But for the stretchable cord, the center of the bar is also changing but in other way, in this case, what will the total kinetic energy look like? Will it be

[tex]K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)[/tex]

But someone tell me it is not necessary to include the term about [tex]x_c[/tex], only the following is enough

[tex]K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2[/tex]

This is very confusing! For rotation, when should we include the translation kinetic energy? and when could we use the rotation kinetic energy only for the whole system?

By the way, if we use a mass instead, what would be different?
 
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  • #2
KFC said:
[tex]K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)[/tex]

Hi KFC! :smile:

You're getting confused about whether I is measured relative to an axis through the ceiling or through the centre of mass.

From the PF Library:
The parallel axis theorem:

The Moment of Inertia of a body about an axis is

[tex]I = (I_C\,+\,md^2)[/tex]

where m is the mass, d is the distance from that axis to the centre of mass, and [itex]I_C[/itex] is the Moment of Inertia about the parallel axis through the centre of mass.

This is the same as your formula if you put x'2 + y'2 = v2 = w2d2. :smile:
But someone tell me it is not necessary to include the term about [tex]x_c[/tex], only the following is enough

[tex]K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2[/tex]

Sorry, don't understand that. :confused:
 
  • #3
Sometimes I think it is useful to think of mv2/2 for each point first.
 

FAQ: Kinetic Energy of Rigid Object Swinging in Vertical Plane

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion.

How is kinetic energy calculated?

Kinetic energy is calculated using the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

How does the kinetic energy of a rigid object swinging in a vertical plane differ from that of a stationary object?

The kinetic energy of a swinging object is constantly changing as it moves through its swinging motion, while the kinetic energy of a stationary object remains constant at zero.

What factors affect the kinetic energy of a swinging object?

The kinetic energy of a swinging object is affected by its mass, velocity, and the height of its swing. The higher the mass and velocity, and the greater the height of the swing, the more kinetic energy the object will have.

How is the kinetic energy of a swinging object converted into other forms of energy?

As the swinging object moves through its vertical plane, its kinetic energy is converted into potential energy at the top of the swing and back into kinetic energy as it swings back down. Some energy may also be lost due to friction and air resistance.

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