Velocity of a Sphere Rolling in a Circular Bowl

[CoffeeCrow]

Homework Statement

A solid sphere of mass M and radius a is released at vertical height y=R and rolls down a circular bowl without slipping, find an expression for the velocity of the sphere's center of mass at the bottom of the bowl.

2. Homework Equations

$I=I_c+Md^2$
$I=\frac {2} {5} Ma^2$
$U=Mgh$
$E_r=\frac {1} {2}I\omega^2$
$E_t=\frac {1} {2}mv^2$

The Attempt at a Solution

The sphere is released from height R, so it's total energy is given by $E=mg(R-a)$ (as the radius of the sphere is non-negligible).
Taking the bottom of the bowl to be the zero of potential energy;

$E_T=mg(R-a)=\frac {1} {2} I\omega^2+\frac {1} {2}mv^2$

At the bottom of the bowl, all gravitational potential energy will have been transformed into translational and rotational kinetic energy.
As the sphere is in rotation around it's contact point, via the parallel axis theorem:
$I=I_c+Md^2$
$I=\frac {2} {5} Ma^2+Ma^2$
$I=\frac {7} {5} Ma^2$

And as $\omega^2=\frac {v^2} {r^2},$

$E_T=mg(R-a)=\frac {1} {2}mv^2+\frac {7} {10} ma^2 (\frac {v^2} {r^2})$

Following this through I come up with $v=\sqrt{\frac {5g(R-a)}{6}}$

Rather than the (correct), $v=\sqrt{\frac {10(R-a)}{7}}$

I know if I only count the rotational kinetic energy I end up with this expression, but I don't understand why I wouldn't also count the translational kinetic energy?

 

[CoffeeCrow]

Sorry, I accidentally hit enter half way through writing this

 

[haruspex]

You've effectively counted the linear KE twice. Either take the sphere's motion as a linear horizontal speed of its mass centre, plus a rotation about the mass centre, or take it as a rotation about the point of contact with the bowl.

 

[CoffeeCrow]

Thank you so much! I think I understand a lot better now (that was a lot more painless than expected).