## kgns Group Title A uniform sphere of mass m and radius r is set free from the top edge of a semicircle halfpipe with radius R. If R > r, what is the time-dependent velocity equation v(t) for the sphere in terms of t, m, r, R and g ignoring any effects of friction? Could anybody help me derive the differential equations? one year ago one year ago

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1. Mashy Group Title

i think you can use conservation of energy principle!!

2. kgns Group Title

I doubt $mgh = \frac{ 1 }{ 2 }mv^2$ will get me something in terms of t

3. MuH4hA Group Title

No, this won't get you somewhere, because the potential gravitational energy will _not_ be "transformed" into kinetic energy to 100%. There will be rotational energy, too.. you have to take that into account..

4. Mashy Group Title

yea.. total kinetic energy = rotational kinectic + translational kinetic.. thats not all that hard cause moment of inertia of the sphere would be known.. and m assuming its without slip!!

5. kgns Group Title

If it were an inclined plane, we could stick with $gtsin \alpha$ with alpha being the angle of inclination. If we consider a semicircle as an infinitesimal sum of inclined planes, we could do $\int\limits_{0}^{t}gsin \alpha(t) d \alpha$ But it's quite a challenge to derive alpha(t). Not even sure if possible

6. MuH4hA Group Title

Shouldn't the acceleration in terms of the angle look like this? |dw:1355060534242:dw| $a(\alpha) = g \, \cos(\alpha)$

7. kgns Group Title

Yeah, but since we want velocity in terms of time, we also need to know alpha in terms of t

8. gleem Group Title

The problem does not specify that the sphere is rolling i.e. not slipping. Assuming it is rolling makes the problem more difficult especially considering the acceleration is not constant.

9. Vincent-Lyon.Fr Group Title

If you write down the equations, you will end up with a non-linear differential equation. There is not classical function that will yield v(t). Besides, if you ignore the effects of friction, then the sphere will not rotate and its kinetic energy will purely be given by the motion of its centre of mass.

10. MuH4hA Group Title

The problem does not specifiy, but the other option doesn't make any sense for this kind of example.