Question

A solid sphere of weight 36.0 N rolls up an incline at an angle of$\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$. At the bottom of the incline the center of mass of the sphere has a translational speed of 4.90 m/s. (a) What is the kinetic energy of the sphere at the bottom of the incline? (b) How far does the sphere travel up along the incline? (c) Does the answer to (b) depend on the sphere’s mass?

Short answer

1. The kinetic energy of the sphere at the bottom of the incline is 61.7 J.
2. Distance that the sphere travel up along the incline is 3.43 m.
3. Answer to (b) does not depend on the sphere’s mass.

Step-by-step solution

Step 1: Given Data

Solid sphere of weight is $36.0\mathrm{N}$,

Incline at an angle is $\mathrm{\theta }=30.0^{°}$,

At the bottom, speed of the center of mass is 4.90 m/s.

Determining the concept

Using the formula${\mathbf{K}}_{\mathbf{rot}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\mathbf{a}$and${\mathbf{K}}_{\mathbf{trans}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$, find the kinetic energy of the sphere at the bottom of the incline and the distance that the sphere travels up along the incline.

Formulae are as follow:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{rot}}=\frac{1}{2}{\mathrm{I\omega }}^2\mathrm{a} \\ {\mathrm{K}}_{\mathrm{trans}}=\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Where, data-custom-editor="chemistry" $\mathrm{\omega }$is angular frequency, m is mass, I is moment of inertia, v is velocity and K is kinetic energy.

(a) Determining the kinetic energy of the sphere at the bottom of the incline

Firstly, for the calculation part, find the mass of the solid sphere,

$\begin{array}{rcl}\mathrm{M}& =& \frac{\mathrm{W}}{\mathrm{g}}\\ & =& \frac{36\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ & =& 3.67\mathrm{kg}\end{array}$

Total kinetic energy can be written as sum of rotational kinetic energy and translational kinetic energy,

${\mathrm{K}}_{\mathrm{total}}=\frac{1}{2}{\mathrm{I}}_{\mathrm{com}}{\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{Mv}}^2$

As,${\mathrm{I}}_{\mathrm{com}}=\frac{2}{5}{\mathrm{MR}}^2$

$$\begin{gathered} \mathrm{K}=\frac{1}{2}\left(\frac{2}{5}{\mathrm{MR}}^2\right){\left(\frac{{\mathrm{v}}_{\mathrm{com}}}{\mathrm{R}}\right)}^2+\frac{1}{2}{\mathrm{Mv}}^2 \\ \mathrm{K}=\frac{7}{10}{\mathrm{Mv}}^2 \\ \mathrm{K}=\frac{7}{10}\left(3.67\mathrm{kg}\right)\left(4.9\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{K}=61.7\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of the sphere at the bottom of the incline is 61.7 J.

(b) Determining the distance that the sphere travels up along the incline

At some height, $\mathrm{h}=\mathrm{d}\mathrm{sin\theta }$, the sphere comes to rest and kinetic energy turns into potential energy $\mathrm{Mgh}$.

Therefore, according to energy conservation law,

$\begin{array}{rcl}\frac{7}{10}{\mathrm{Mv}}_{\mathrm{com}}^2& =& \mathrm{Mgd}\mathrm{sin\theta }\\ \mathrm{d}& =& \frac{7{\mathrm{v}}_{\mathrm{com}}^2}{10\mathrm{g}\mathrm{sin\theta }}\\ \mathrm{d}& =& \frac{7\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)}{10\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin 30.0^{°}}\\ \mathrm{d}& =& 3.43\mathrm{m}\end{array}$

Hence, distance that the sphere travel up along the incline is 3.43 m.

(c) Determining if the answer to (b) depend on the sphere’s mass

In part (b), for the calculation of d, M gets cancelled out. As the answer is independent of mass, it is also independent of the sphere’s weight.

Hence,the answer to (b) does not depend on the sphere’s mass.

Using the formula for rotational kinetic energy and translational kinetic energy along with the conservation of energy, the total energy of the rolling sphere can be found. Using this total energy and law of conservation of energy, it is possible to find the height that would be achieved by the sphere on the incline if the angle is known.