Particle sliding down a frictionless sphere

[attilathedud]

Homework Statement

A particle of mass m slides down a fixed frictionless sphere of radius R starting from rest at the top.
a. In terms of m, g, R, and Θ, determine each of the following for the particle while it is sliding on the sphere.
i. The kinetic energy of the particle.
ii. The centripetal acceleration of the mass.
iii. The tangential acceleration of the mass.
b. Determine the value of Θ at which the particle leaves the sphere.

Homework Equations

None given.

The Attempt at a Solution

a.
i.
Ki + Vi = Kf + Vf
0 + mgy = 1/2mv^2 + 0
gy = 1/2v^2
g(R/2) = v^2/2
v = sqrt(gR)

ii.
ar = v^2 / R
ar = gR/R
ar = g

iii.
v = r(dΘ / dt)
sqrt(gR) = R(Θ/t)
gR = (R(Θ / t)) ^ 2
t = d/v
gR = (R(Θ / d / sqrt(gR))^2
gR = R(Θsqrt(gR)/d)^2

b. Pretty sure the fact I massively screwed up part iii is making this section impossible. And having the centripetal acceleration equal gravity seems incorrect as well.

Edit: Seems like I wasn't the first one to post this. My apologies.

The forum adds the template tags to the bottom of the current post every time you hit the "Preview Post" button...

 

[vela]

i.
Ki + Vi = Kf + Vf
0 + mgy = 1/2mv^2 + 0
gy = 1/2v^2
g(R/2) = v^2/2
v = sqrt(gR)

Using energy is the right approach, but you need to find an expression for $v$ as a function of $\theta$. Don't set $y$ to a specific value as you did here. Instead, express it as a function of $R$ and $\theta$.

ii.
ar = v^2 / R
ar = gR/R
ar = g

This part will work out once you fix part i.

iii.
v = r(dΘ / dt)
sqrt(gR) = R(Θ/t)
gR = (R(Θ / t)) ^ 2
t = d/v
gR = (R(Θ / d / sqrt(gR))^2
gR = R(Θsqrt(gR)/d)^2

It would be simpler to resolve the weight of the particle into radial and tangential components.

b. Pretty sure the fact I massively screwed up part iii is making this section impossible. And having the centripetal acceleration equal gravity seems incorrect as well.

How would you interpret "particle leaves sphere" in terms of the forces acting on the particle?