Potential Energy of Oscillating surface

[catmando]

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))

 

[tiny-tim]

hi catmando!

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 ,

(perhaps you could mention O and OG?)

but your answer is certainly correct

(except for a minus sign )

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