# Circular motion, find speed

## Homework Statement

A particle is at rest at the apex A of a smooth fixed hemisphere whose base is horizontal. The hemisphere has centre O and radius a. The particle is then displaced very slightly from rest and moves on the surface of the hemisphere. At the point P on the surface where angle AOP = α the particle has speed v. Find an expression for v in terms of a, g and α.

## The Attempt at a Solution

So I’ve worked like this:
Total energy at A = PE + KE = amg + 0
Total energy at P = PE + KE = (0.5m(v^2)) + xmg
x = a – y
(cos α)/y = (sin90)/a => y = a(cos α)
=> x = a – (a(cos α)) = a (1 - cos α)
=> Total energy at P = PE + KE = (0.5m(v^2)) + amg (1 - cos α)

Therefore (Total energy at A) = (Total energy at P) gives
amg = (0.5m(v^2)) + amg (1 - cos α)
ag = 0.5(v^2) + ag (1 – cos α)
ag – ag (1 – cos α) = 0.5(v^2)
2ag (1 – 1 + cos α) = v^2
v = sqrt (2ag (cos α))

However, the correct answer is v = sqrt (2ag (1 - cos α))

Where’s the problem?

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## The Attempt at a Solution

So I’ve worked like this:
Total energy at A = PE + KE = amg + 0
Total energy at P = PE + KE = (0.5m(v^2)) + xmg
You calculated the initial potential energy with respect to the base of the semiphere. PE=mg*(height above the base). The height is y in your picture.

ehild

thanks