Question

It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?Part AA uniform solid cylinder.Part BA uniform sphere.Part CA thin-walled hollow sphere.Part DA hollow, cylinder with outer radius R and inner radius R/2.

Answer 1

Answer:

Part a)

$f = \frac{0.5}{1.5} = \frac{1}{3}$

B) uniform Sphere

$f = \frac{2}{7}$

C) uniform hollow sphere

$f = \frac{2}{5}$

D) uniform hollow cylinder with inner radius R/2 and outer radius R

$f = \frac{5}{13}$

Explanation:

As we know that fraction of total energy as rotational energy is given as

$f = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2}$

now we have

$f = \frac{mk^2(\frac{v^2}{R^2})}{mv^2 + mk^2(\frac{v^2}{R^2})}$

$f = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}$

now we have

A) uniform Solid cylinder

for cylinder we know that

$\frac{k^2}{R^2} = 0.5$

$f = \frac{0.5}{1.5} = \frac{1}{3}$

B) uniform Sphere

for sphere we know that

$\frac{k^2}{R^2} = \frac{2}{5}$

$f = \frac{0.4}{1.4} = \frac{2}{7}$

C) uniform hollow sphere

for hollow sphere we know that

$\frac{k^2}{R^2} = \frac{2}{3}$

$f = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}$

D) uniform hollow cylinder with inner radius R/2 and outer radius R

for annular cylinder we know that

$\frac{k^2}{R^2} = \frac{5}{8}$

$f = \frac{\frac{5}{8}}{\frac{13}{8}} = \frac{5}{13}$

Answer 2

The fraction of the total kinetic energy that is rotational for a uniform sphere is $\frac{1}{3}$

How to calculate rotational kinetic energy.

Mathematically, the rotational kinetic energy of an object is given by this formula:

$K.E_{rot}=\frac{1}{2} I\omega^2$  

Where:

• I is the moment of inertia.

• $\omega$ angular velocity.

Since we know that half of the total kinetic energy for a hollow, cylindrical shell that is rolling without slipping on a horizontal surface is translational and the other half is rotational. Thus, this fraction is given by this mathematical expression:

$K.E=\frac{\frac{k^2}{R^2} }{1+\frac{k^2}{R^2}}$

a. For a uniform sphere:

$\frac{k^2}{R^2}=0.5$

Substituting, we have:

$K.E=\frac{0.5 }{1+0.5}\\\\K.E=\frac{0.5 }{1.5}\\\\K.E =\frac{1}{3}$

b. For a thin-walled hollow sphere:

$\frac{k^2}{R^2}=\frac{2}{5}$

Substituting, we have:

$K.E=\frac{\frac{2}{5} }{1+\frac{2}{5}}\\\\K.E=\frac{0.4 }{1.4}\\\\K.E =\frac{2}{7}$

c. For a uniform hollow sphere:

$\frac{k^2}{R^2}=\frac{2}{3}$

Substituting, we have:

$K.E=\frac{\frac{2}{3} }{1+\frac{2}{3}}\\\\K.E=\frac{0.7 }{1.7}\\\\K.E =\frac{2}{5}$

d. For a hollow sphere with outer radius (R) and inner radius:

$\frac{k^2}{R^2}=\frac{5}{8}$

Substituting, we have:

$K.E=\frac{\frac{5}{8} }{1+\frac{5}{8}}\\\\K.E=\frac{0.625 }{1.625}\\\\K.E =\frac{5}{13}$

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