Velocity of a Sphere Rolling in a Circular Bowl

In summary, the problem involves a solid sphere rolling down a circular bowl without slipping. The velocity of the sphere's center of mass at the bottom of the bowl can be found by considering the total energy of the sphere, which is equal to the sum of its rotational and translational kinetic energy. By applying the parallel axis theorem, the moment of inertia of the sphere can be calculated. However, when calculating the velocity, the translational kinetic energy should not be counted twice, as it is already included in the rotational kinetic energy.
  • #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


##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]

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;


##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:
[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?
 
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  • #2
Sorry, I accidentally hit enter half way through writing this
 
  • #3
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.
 
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  • #4
Thank you so much! I think I understand a lot better now (that was a lot more painless than expected).
 

FAQ: Velocity of a Sphere Rolling in a Circular Bowl

1. What is the velocity of a sphere rolling in a circular bowl?

The velocity of a sphere rolling in a circular bowl depends on factors such as the size and weight of the sphere, the slope and material of the bowl, and any external forces acting on the sphere. It can be calculated using the formula v = rω, where v is the velocity, r is the radius of the bowl, and ω is the angular velocity of the sphere.

2. How does the shape of the bowl affect the velocity of the sphere?

The shape of the bowl can greatly affect the velocity of the sphere. A deeper bowl with steeper sides will have a higher velocity, as gravity will pull the sphere down at a faster rate. A shallower bowl with gentler sides will have a lower velocity, as the sphere will roll more slowly due to less gravitational pull.

3. Does the weight of the sphere affect its velocity in a circular bowl?

Yes, the weight of the sphere can affect its velocity in a circular bowl. A heavier sphere will have a higher velocity, as it will have more momentum and will be more resistant to changes in speed. A lighter sphere will have a lower velocity, as it will have less momentum and will be easier to change in speed.

4. How does friction play a role in the velocity of a sphere rolling in a circular bowl?

Friction can greatly affect the velocity of a sphere rolling in a circular bowl. If there is high friction between the sphere and the bowl, the velocity will be lower as the sphere will experience more resistance while rolling. If there is low friction, the velocity will be higher as the sphere can roll more easily without much resistance.

5. Can external forces affect the velocity of a sphere rolling in a circular bowl?

Yes, external forces such as wind or pushing the sphere can affect its velocity in a circular bowl. If the external force is in the same direction as the motion of the sphere, it can increase the velocity. If the external force is in the opposite direction, it can decrease the velocity. However, in an ideal scenario, without any external forces, the velocity of the sphere will remain constant due to the conservation of energy.

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