- #1

CoffeeCrow

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## 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

2. Homework Equations

##I=I_c+Md^2##

**[itex]I=\frac {2} {5} Ma^2[/itex]**

##U=Mgh##

[itex]E_r=\frac {1} {2}I\omega^2[/itex]

[itex]E_t=\frac {1} {2}mv^2[/itex]

##U=Mgh##

[itex]E_r=\frac {1} {2}I\omega^2[/itex]

[itex]E_t=\frac {1} {2}mv^2[/itex]

## The Attempt at a Solution

**The sphere is released from height R, so it's total energy is given by [itex]E=mg(R-a)[/itex] (as the radius of the sphere is non-negligible).**

Taking the bottom of the bowl to be the zero of potential energy;

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##

As the sphere is in rotation around it's contact point, via the parallel axis theorem:

[itex] I=I_c+Md^2 [/itex]

## I=\frac {2} {5} Ma^2+Ma^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:

[itex] I=I_c+Md^2 [/itex]

## 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?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?

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