Potential Energy of Oscillating surface

In summary, the rectangular sheet of thin material is rolled into a portion of a circular cylindrical surface, resting under the action of gravity on a fixed cylindrical surface. When rolled slightly to the left or right, it will rock from side to side without slipping. The expression for the gravitational potential energy of the object at any instant during the oscillation is PE = mg(kr + kr cos (β + θ) - (R+r) + (R+r)cos(β)), where β is the angle between the fixed and rolled surfaces, θ is the angle between the vertical centre line and the fixed surface, and k is a constant.
  • #1
catmando
7
0

Homework Statement



A rectangular sheet of thin material is rolled into a portion of a circular cylindrical surface of radius r. It rests under the action of gravity on a fixed cylindrical surface of radius R, symmetrical about the vertical centre line as shown below. If rolled slightly to the left or the right and released, the object will rock from side to side (assume no slipping).

write an expression for the gravitational potential energy of the object at some instant during the oscillation. Use the rest position of the centre of mass as the potential energy datum

ht tp://s11.postimage.org/gdrfgsoep/lol.jpg

b = k*r

Homework Equations



PE = mgh

The Attempt at a Solution



h = b - x

x = (R+r) - (R+r)cos(β)
b = kr - a
a = kr cos (β + θ)

h = kr +kr cos (β + θ) -(R+r) + (R+r)cos(β)

therefore potential energy

PE = mg(kr +kr cos (β + θ) -(R+r) + (R+r)cos(β))
 
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  • #2
hi catmando! :smile:
catmando said:
h = b - x

x = (R+r) - (R+r)cos(β)
b = kr - a
a = kr cos (β + θ)

h = kr +kr cos (β + θ) -(R+r) + (R+r)cos(β)

therefore potential energy

PE = mg(kr +kr cos (β + θ) -(R+r) + (R+r)cos(β))

i don't follow your reasoning :redface:,

(perhaps you could mention O and OG?)

but your answer is certainly correct :smile:

(except for a minus sign :wink:)

to finish it, what is θ as a function of β ?​
 

1. What is potential energy of oscillating surface?

Potential energy of oscillating surface refers to the energy that is stored within a surface or system when it is in a state of oscillation or vibration. It is a type of energy that is associated with the displacement of an object from its equilibrium position.

2. How is potential energy of oscillating surface calculated?

The potential energy of an oscillating surface can be calculated using the equation PE = 1/2kx², where PE is the potential energy, k is the spring constant, and x is the displacement of the surface from its equilibrium position. This equation is derived from Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.

3. What factors affect the potential energy of an oscillating surface?

The potential energy of an oscillating surface is affected by two main factors: the amplitude of the oscillation and the spring constant. The greater the amplitude of the oscillation, the higher the potential energy will be. Similarly, a higher spring constant will result in a greater potential energy for the same displacement.

4. How does potential energy of oscillating surface relate to kinetic energy?

Potential energy and kinetic energy are two forms of energy that are interrelated. In the case of an oscillating surface, potential energy is at its maximum when the surface is at its equilibrium position, and kinetic energy is at its maximum when the surface is at its maximum displacement from the equilibrium position. As the surface oscillates, potential energy is converted into kinetic energy and vice versa.

5. What is the significance of potential energy of oscillating surface?

The potential energy of oscillating surface plays a crucial role in understanding the behavior of oscillating systems, such as springs and pendulums. It helps us to analyze and predict the motion of these systems, and is also used in various applications such as in the design of shock absorbers and energy harvesting devices.

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