A small solid sphere, with radius 0.25 cm and mass 0.70 g rolls without slipping

Question

A small solid sphere, with radius 0.25 cm and mass 0.70 g rolls without slipping on the inside of a large fixed hemisphere with radius 23 cm and a vertical axis of symmetry. The sphere starts at the top from rest.
(a) What is its kinetic energy at the bottom?
J

(b) What fraction of its kinetic energy at the bottom is associated with rotation about an axis through its center of mass?
KE

(c) What is the magnitude of the normal force on the hemisphere from the sphere when the sphere reaches the bottom?
N

Explanation / Answer

(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:

KE = mgR = (.70 g * 1 kg/1000 g)(9.8 m/s²)(23 cm * 1 m/100 cm)

KE = .001579 J

(b) The kinetic energy associated with rotation is given as:

KER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:

v = w * r

Substituting this into the expression for the rotational kinetic energy:

KER = (1/2) (I / r²) v²

Note that the expression for the translational kinetic energy is:

KET = (1/2) m v²

Thus, the total kinetic energy of the ball at the base of the hemisphere is;

KE = KER + KET = (v²/2)( I/r² + m)

The fraction of the rotational kinetic energy over the total is:

f = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)

f = ( I/r²)/( I/r² + m)

Finally, the moment of inertia, I, of a sphere is:

I = (2/5)mr²

Substituting this into the expression for f:

f = (2/5)m/( (2/5)m + m)

f = (2/5)/( (2/5) + 1)

f = 2/7

(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:

N = Fgrav + Fcent

N = mg + mv²/R

To find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:

(1/2) (I / r) v² = (2/7)(mgR)

(1/2) ((2/5)m) v² = (2/7)(mgR)

(1/5) v² = (2/7)(gR)

v = sqrt((10/7)(gR))

v = sqrt((10/7)(9.8 * 23 cm * 1m/100 cm))

v = sqrt((10/7)(9.8 * 23 cm * 1m/100 cm))

v = 1.795 m/s

Finally, the normal force is computed as:

N = mg + mv²/R

N = (.70 g)(9.81 m/s²) + (.70 g)(1.795 m/s)²/(23 cm)

N = .01667 Newtons

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