Question

A sphere of radius$r=\mathbf{34}\mathbf{.5}\mathbf{c}\mathbf{m}$and mass$m=1.80kg$starts from rest and rolls without slipping down a 30.0° incline that is 10.0 m long. (a) Calculate its translational and rotational speeds when it reaches the bottom. (b) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: (c) do your answers in (a) and (b) depend on the radius of the sphere or its mass?

Short answer

1. The translational speed of a sphere is 8.37 m/s. The rotational speed of a sphere is$24.261\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}$.
2. The ratio of translational energy to rotational kinetic energy at the bottom is 5/2.
3. In part (a), only the rotational speed depends on the sphere's radius and both speeds are independent of the mass. Part (b) is independent of the mass and radius of the sphere.

Step-by-step solution

Identification of given data 

The given data can be listed below as:

• The radius of the sphere is$r=34.5\mathrm{c}\mathrm{m}\left(\frac{1\mathrm{m}}{100\mathrm{c}\mathrm{m}}\right)=0.345\mathrm{m}$.
• The mass of a sphere is$m=1.80\mathrm{k}\mathrm{g}$.
• The angle of inclination is$\theta ={30.0}^{\circ }$.
• The length of the inclined plane is $l=10.0\mathrm{m}$.
• The acceleration due to gravity is$g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Understanding the motion of the sphere

The sphere rolls down from the top of an inclined plane. At the top, it has potential energy. During the motion from an inclined plane, its energy is converted into kinetic energy. So, at the bottom, it only has kinetic energy.The total energy of the sphere at the bottom is the sum of rotational energy and translational kinetic energy.

Determination of the total kinetic energy and the potential energy of the sphere

The sphere is at rest at the top of the incline. The rotational kinetic energy of the sphere can be expressed as:

$K_r=\frac{1}{2}I{\omega }^2$ … (i)

Here,$\omega$is the angular speed of the sphere.

The moment of inertia of the sphere can be expressed as:

$I=\frac{2}{5}m{r}^2$

Substitute the values in equation (i).

$\begin{array}{rl}{K}_r& =\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\omega }^2\\ K_r& =\frac{1}{5}m{r}^2{\omega }^2\end{array}$

$K_r=\frac{1}{5}m{v}^2$ … (ii)

Here,$v$is the translational speed of the sphere.

The translational kinetic energy of the sphere can be expressed as:

$K_t=\frac{1}{2}m{v}^2$ … (iii)

The total kinetic energy is equal to the sum of the translational and rotational kinetic energies. Add equations (ii) and (iii) to obtain the total kinetic energy.

$\begin{array}{rl}{K}_T& =K_r+K_t\\ K_T& =\frac{1}{5}m{v}^2+\frac{1}{2}m{v}^2\end{array}$

$K_T=\frac{7}{10}m{v}^2$ … (iv)

The potential energy of the sphere can be expressed as:

$P.E=mgh$ … (v)

The height of the inclined plane can be expressed as:

$h=l\sin \theta$

Substitute the values in equation (v).

$P.E=mgl\sin \theta$ … (vi)

(a) Determination of the translational speed of the sphere

The potential energy gets converted into kinetic energy and is equal to the total kinetic energy. Equate equations (iv) and (vi). This can be expressed as:

$\begin{array}{rl}P.E& =K_T\\ mgl\sin \theta & =\frac{7}{10}m{v}^2\\ v^2& =\frac{10mgl\sin \theta }{7m}\\ v& =\sqrt{\frac{10gl\sin \theta }{7}}\end{array}$

Substitute the values in the above equation.

$\begin{array}{rl}v& =\sqrt{\frac{10\times 9.81\mathrm{m}/{\mathrm{s}}^2\times 10\mathrm{m}\times \sin {30}^{\circ }}{7}}\\ & =\sqrt{70.07}\mathrm{m}/\mathrm{s}\\ & =8.37\mathrm{m}/\mathrm{s}\end{array}$

Thus, the translational speed of a sphere is 8.37 m/s.

Determination of the rotational speed of the sphere

The rotational speed of the sphere can be expressed as:

$\omega =\frac{v}{r}$

Substitute the values in the above equation.

$\begin{array}{rl}\omega & =\frac{8.37\mathrm{m}/\mathrm{s}}{0.345\mathrm{m}}\\ & =24.26\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\end{array}$

Thus, the rotational speed of a sphere is $24.26\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}$.

(b) Determination of the ratio of translational to the rotational kinetic energies of the sphere

From equations (ii) and (iii), the ratio of translational to rotational kinetic energies can be expressed as:

$\begin{array}{rl}\frac{K_t}{K_r}& =\frac{\frac{1}{2}m{v}^2}{\frac{1}{5}m{v}^2}\\ \frac{K_t}{K_r}& =\frac{5}{2}\end{array}$

Thus, the ratio of translational to rotational kinetic energies at the bottom is 5/2.

(c) Analysis of the dependency of the results of (a) and (b) on the radius and mass of the sphere

From part (a), only the rotational speed depends on the radius. The translational speed of the sphere is independent of the radius and mass. None of the speeds depends on the mass of the sphere.

From part (b), the ratio of translational and rotational kinetic energies is independent of the mass and radius of the sphere.

Thus, in part (a), only the rotational speed depends on the sphere's radius and both speeds are independent of the mass. Part (b) is independent of the mass and radius of the sphere.