- #1
ScienceIsMyLady
- 11
- 1
- Homework Statement
- Calculating when does a particle falls off a dome
- Relevant Equations
- a = v^2 / r
A particle was resting on top of a dome and given a negligible push such that it falls. The question is at what angle will the particle fall off the dome.
The solution is that,
[tex]m\frac{v^2}{a} = mg\cos\theta - R[/tex]
and by conservation of energy
[tex]0 + mga = \frac{1}{2}mv^2 + mga\cos\theta[/tex]
then the angle is found to be
[tex]\theta=\cos^{-1}\frac{2}{3}[/tex]
What I don't understand is, when the particle exerts force on the dome due to gravity, why doesn't the dome follows Newton's third law and exert an equal but opposite force on the particle
[tex]mg\cos\theta = R[/tex]
but exerts a smaller force back to the particle such that a net acceleration towards the center is resulted in the particle instead?