Conservation of energy in rotating bodies

In summary, the conservation of energy equation is basically GPE is converted to KE of block and KE of cylinder. The KE of the cylinder is 1/2mv^2, where m is its mass and v is the velocity of its COM (which is the centre of cylinder). However, I viewed the cylinder as rotating about the point of contact with the step, and thus thought that its KE was 1/2Iw^2, where I = 3/2mr^2 (by parallel axis theorem), and rw = v, where v is the velocity of the centre of mass.
  • #1
phantomvommand
242
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Screenshot 2021-03-01 at 2.29.09 AM.png


The conservation of energy equation is basically GPE is converted to KE of block and KE of cylinder.

To get the correct answer, the KE of the cylinder is 1/2mv^2, where m is its mass and v is the velocity of its COM (which is the centre of cylinder).
However, I viewed the cylinder as rotating about the point of contact with the step, and thus thought that its KE was 1/2Iw^2, where I = 3/2mr^2 (by parallel axis theorem), and rw = v, where v is the velocity of the centre of mass.

Why is my calculation of KE for the cylinder wrong?

Thank you for your help.

This problem can be found in Jaan Kalda's Mechanics problems, or in SS Krotov's Aptitude Test Problems.
 
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  • #2
phantomvommand said:
Summary:: Why is the kinetic energy I ascribe to the cylinder in the following question wrong?

Why is my calculation of KE for the cylinder wrong?
Because the cylinder is not rotating.
 
  • #3
phantomvommand said:
I viewed the cylinder as rotating about the point of contact with the step
Why about this contact and not the other one? Why rotating at all?
 
  • #4
as they above say your work is wrong, but u have some ok intuition. cylinder not rotate about its mass centre, however the mass centre point do rotate in circular arc about the contact point on right (well, so long as it is still contact!). this you can use to advantage! say, mass centre of cylinder has rotate by ##\theta## about right point of contact. let ##v = |\underline{v}|## be speed of the mass centre. can first use the energy conserve$$2gr(1-\cos{\theta}) = v^2$$now, remember that mass centre move as if it were particle which is act upon by resultant force, so you now pretend mass centre is a particle, and resolve resultant of force toward right contact point (i.e. you calculate centripetal component):$$-N + mg\cos{\theta} + N_2 \sin{2\theta} = \frac{mv^2}{r}$$where ##N## is force of contact from right point, and ##N_2## force of contact from left point. now you must do examine cases when ##N = 0## and ##N_2 = 0## respective, i.e. set each to ##0## in turn and use two equation to eliminate ##v##, and find ##\theta_{\text{max}}## for each. then you can tell, which lose contact first.

see - much more easy than do use lot of messy coordinate and difficult equation, yes ;)
 

What is conservation of energy in rotating bodies?

Conservation of energy in rotating bodies is a fundamental physical law that states that the total energy of a system remains constant over time. In other words, energy cannot be created or destroyed, but can only be transferred from one form to another.

How does conservation of energy apply to rotating bodies?

In rotating bodies, conservation of energy means that the total energy of the system, which includes both kinetic energy (energy of motion) and potential energy (energy of position), remains constant as the body rotates. This means that as the body rotates, the distribution of energy between kinetic and potential energy may change, but the total amount of energy remains the same.

What is the equation for calculating the total energy of a rotating body?

The equation for calculating the total energy of a rotating body is E = 1/2 * I * ω^2, where E is the total energy, I is the moment of inertia of the body, and ω is the angular velocity (rate of rotation) of the body.

How does conservation of energy in rotating bodies relate to the law of conservation of angular momentum?

The law of conservation of angular momentum states that the total angular momentum of a system remains constant in the absence of external torques. This law is closely related to conservation of energy in rotating bodies, as both laws involve the conservation of a physical quantity (energy or angular momentum) in a closed system.

What are some real-life applications of conservation of energy in rotating bodies?

Conservation of energy in rotating bodies has many practical applications, including in the design and operation of machines and equipment such as turbines, motors, and flywheels. It is also important in understanding the behavior of celestial bodies, such as planets and stars, which rotate and have their own unique energy distribution.

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