# Increased normal force due to rotation

1. Jul 7, 2004

### quarkman

I am studying for my graduate exam in physics and came across a stange phenomenon and I would like to make sure I am reasoning it correctly:

Start with a hemispherical bowl and place a small sphere (relatively) at the edge. Let the sphere roll towards the center of the bowl, thus gaining kinetic energy. At the bottom of the bowl, I have calculated the normal force the small sphere exerts on the hemispherical bowl, allowing the radius of the sphere be much less than the radius of the bowl. This force is much greater (by 17/7) than just (mass sphere) x (gravitational acceleration)!

Through my calculations I have reasoned the sphere has both translational and kinetic energy at the bottom of the bowl. From the angular speed of the sphere, I have calculated the required radial acceleration of a point on the edge of the sphere. Now I let the radius of the sphere become very small compared to the radius of the bowl it is in. Now there is a force on the bowl due to this acceleration of the sphere. From Newton's second law I finally determine that the normal force the sphere exerts on the bowl is the sum of the gravitatonal force on the sphere plus this radial acceleration due to the rotation of the sphere:
i.e. Fext = N - mg = ma
with:
a = radial acceleration of sphere
N = normal force
g = gravitational acceleration
m = mass of sphere

This analysis appears sketchy to me and I may not be explaining myself correctly, but I get the answer I am supposed to get for the normal force. Could anyone help clarify this for me? Thanks.

2. Jul 7, 2004

### arildno

Perhaps you think this analysis is clearer:
1. Rolling condition:
This means that the sphere's contact point's velocity is zero (i.e., equal to the velocity of the bowl's contact point)
Two important implications:
a) The frictional force acting on the contact point is static, hence, it does no work, and the mechanical energy of the system is conserved.
b)The angular velocity is related to the velocity of C.M (assuming C.M to be the center of the sphere) by:
$$\omega=\frac{v_{c.m}}{r_{s}}$$

2. Energy conservation:
We place the sphere initially at the rim of the bowl, so that it's potential value relative to the bottom of the bowl is mgR
The sphere's potential value when it is at the bottom is $$mgr_{s}$$
Hence, we have:
$$\frac{m}{2}v^{2}_{c.m}+\frac{m}{2}(\frac{2}{5})v_{c.m}^{2}=mg(R-r_{s})$$
Or:
$$v_{c.m}^{2}=\frac{10}{7}g(R-r_{s})$$

3. Newton's 2.law:
Since the C.M is traversing a circle, the net sum of forces must give the centripetal acceleration of C.M:
$$N-mg=m\frac{v_{c.m}^{2}}{R-r_{s}}$$
Or, as you found:
$$N=\frac{17}{7}mg$$

3. Jul 7, 2004

### quarkman

Great answer. You have clarified things very well. Thank you very much for your time with this problem. I am interested in any speculation you may have for the reasons why the original problem would have told me to assume a small sphere radius. Was this information there to lead me astray, or is there some implication when the radius of the sphere approaches the radius of the bowl? Thanks again for your help.

4. Jul 7, 2004

### arildno

I think it was included to make the problem "easier"
For example, some might have forgotten that the centripetal acceleration is calculated with respect to the actual radius, R-r; others might forget that the potential difference is mg(R-r), and so on..

Welcome to PF, BTW.

5. Jul 17, 2004

### rayjohn01

To AR
Not querying your maths , but I'm having a puzzling time thinking about a sphere which is as large as the bowl (i.e. it just fits inside) --can you resolve this ?

6. Jul 17, 2004

### arildno

rayjohn:
In that case, the initial potential energy will be equal to the "final" potential energy, which means that the kinetic energies must be equal as well (i.e, zero, since the initial kinetic energy is given as zero)

7. Jul 17, 2004

### rayjohn01

To AR
it's a quibble but it appears to depend on the exact starting assumptions -- if the ball starts on top of the lip it has to tip-in. If it starts inside but central to the lip it will fall in directly, if it's top is at the lip it cannot move hence Fn = mg. his comes as a result of having a finite ball without stating the precise initial conditions.

8. Jul 17, 2004

### arildno

Certainly, rayjohn:
But I solved it explicitly under the assumption that the C.M. of the sphere is only horizontally displaced from the contact point on the rim.
It is certainly true that the initial potential energy will be different if we allow the contact point on the edge to be in contact with a contact point on the sphere that does not have a horizontal normal; however, in that case, there will be an initial, highly problematic phase where energy conservation only can be assumed if the sphere can rotate on the sharp edge with no contact point velocity.
In my view, it is more probable that in such a case, there will be an initial slip; i.e, either loss of contact with the bowl's surface, or a (disssipative) sliding phase along the bowl's surface.
Either of these scenarios would bring in complications not envisaged by the problem maker; in particular, one ought to assume that the accuracy level of simple energy maths is rather poor.