Question

A hollow spherical shell with mass 2.50 kg rolls without slipping down a slope that makes an angle of 36.0degrees with the horizontal. Find the magnitude of the acceleration acm of the center of mass of the spherical shell?Take the free-fall acceleration to be g = 9.80 m/s^2, then Find the magnitude of the frictional force acting on the spherical shell.Take the free-fall acceleration to be g = 9.80 m/s^2.

Answer 1

Concepts and reason

The concepts required to solve this problem are torque, and Newton’s second law.

Initially use the parallel axis theorem to solve for the moment of inertia of the hollow spherical shell. Then use the net torque equation and rearrange it to solve for acceleration of the center of mass.

Later, use Newton’s second law to solve for the friction force acting on the spherical shell. Then use the Newton’s second to solve for normal force.

Finally rearrange the friction force formula to solve for coefficient of friction.

Fundamentals

The torque is given as,

$\tau = rF$

Here, F is the force on the object, and r is the perpendicular distance from the axis of rotation.

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The equation of the Newton’s second law is,

$\sum {F = ma}$

Here, $\sum F$ is the net force on the object, $m$ is mass of the object, and $a$ is the acceleration of the object.

The force of gravity on any object parallel to the incline plane is,

${F_{{\rm{g}}\parallel }} = mg\sin \theta$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle of inclination.

The force of gravity on any object perpendicular to the incline plane is,

${F_{{\rm{g}} \bot }} = mg\cos \theta$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle of inclination.

The net torque on an object is,

$\sum \tau = I\alpha$

Here, $\sum \tau$ is the sum of all the torques on the object.

The angular acceleration in terms of linear acceleration is,

$\alpha = \frac{{{a_{{\rm{cm}}}}}}{r}$

Here, ${a_{{\rm{cm}}}}$ is the acceleration of the center of mass of the body, and r is the radius of rotation.

The force of friction is given as,

$f = \mu {F_{\rm{N}}}$

Here, $\mu$ is the coefficient of friction, and ${F_{\rm{N}}}$ is the normal force.

According to the parallel axis theorem, the moment of inertia of the rotating body about any axis parallel to axis passing through its center is equal to the sum of moment of inertia about the axis passing through its center and product of mass and square of distance of the parallel axis from the axis passing from the center that is,

$I = {I_{{\rm{cm}}}} + m{r^2}$

Here, $I$ is the moment of inertia about any axis, ${I_{{\rm{cm}}}}$ is the moment of inertia about the axis passing from the center of the body, $m$ is the mass of the body, and $r$ is the distance of parallel axis from the axis passing from the center.

The moment of inertia of a hollow sphere about an axis passing through its center is,

${I_{{\rm{cm}}}} = \frac{2}{3}m{r^2}$

Here, $m$ is the mass of the hollow sphere, and $r$ is the radius of hollow sphere.

Sign convention used is as follows:

All the forces downward are negative and upward are positive.

Use the parallel axis theorem.

Substitute $\frac{2}{3}m{r^2}$ for ${I_{{\rm{cm}}}}$ in the equation $I = {I_{{\rm{cm}}}} + m{r^2}$.

$\begin{array}{c}\\I = \frac{2}{3}m{r^2} + m{r^2}\\\\ = \frac{5}{3}m{r^2}\\\end{array}$

Substitute $rF$ for $\sum \tau$, $\frac{5}{3}m{r^2}$ for $I$, and $\frac{{{a_{{\rm{cm}}}}}}{r}$ for $\alpha$ in the torque equation $\sum \tau = I\alpha$.

$rF = \left( {\frac{5}{3}m{r^2}} \right)\left( {\frac{{{a_{{\rm{cm}}}}}}{r}} \right)$

Substitute $mg\sin \theta$ for $F$ in the above equation $rF = \left( {\frac{5}{3}m{r^2}} \right)\left( {\frac{{{a_{{\rm{cm}}}}}}{r}} \right)$ and solve for acceleration.

$\begin{array}{c}\\r\left( {mg\sin \theta } \right) = \left( {\frac{5}{3}m{r^2}} \right)\left( {\frac{{{a_{{\rm{cm}}}}}}{r}} \right)\\\\{a_{{\rm{cm}}}} = \frac{3}{5}g\sin \theta \\\end{array}$

Substitute $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$, and $36^\circ$ for $\theta$ in the equation ${a_{{\rm{cm}}}} = \frac{3}{5}g\sin \theta$.

$\begin{array}{c}\\{a_{{\rm{cm}}}} = \frac{3}{5}\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 36^\circ \\\\ = 3.46{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Substitute ${F_{{\rm{g}}\parallel }} - f$ for $\sum F$ and ${a_{{\rm{cm}}}}$ for $a$ in the equation $\sum F = ma$.

${F_{{\rm{g}}\parallel }} - f = m{a_{{\rm{cm}}}}$

Substitute $mg\sin \theta$ for ${F_{{\rm{g}}\parallel }}$ in the equation ${F_{{\rm{g}}\parallel }} - f = m{a_{{\rm{cm}}}}$ and solve for the friction.

$\begin{array}{c}\\mg\sin \theta - f = m{a_{{\rm{cm}}}}\\\\f = mg\sin \theta - m{a_{{\rm{cm}}}}\\\\ = m\left( {g\sin \theta - {a_{{\rm{cm}}}}} \right)\\\end{array}$

Substitute $2.30{\rm{ kg}}$ for $m$, $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$, $36^\circ$ for $\theta$, and $3.46{\rm{ m/}}{{\rm{s}}^2}$ for ${a_{{\rm{cm}}}}$ in the equation $f = m\left( {g\sin \theta - {a_{{\rm{cm}}}}} \right)$.

$\begin{array}{c}\\f = \left( {2.30{\rm{ kg}}} \right)\left( {\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 36^\circ - 3.46{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 5.3{\rm{ N}}\\\end{array}$

Use the Newton’s second law perpendicular to the incline.

Substitute ${F_{\rm{N}}} - {F_{{\rm{g}} \bot }}$ for $\sum F$ in the equation $\sum F = ma$.

${F_{\rm{N}}} - {F_{{\rm{g}} \bot }} = ma$

Substitute $0$ for $a$ and $mg\cos \theta$ for ${F_{{\rm{g}} \bot }}$ in the above equation ${F_{\rm{N}}} - {F_{{\rm{g}} \bot }} = ma$ and solve for the normal force.

$\begin{array}{c}\\{F_{\rm{N}}} - mg\cos \theta = 0\\\\{F_{\rm{N}}} = mg\cos \theta \\\end{array}$

Use the friction force formula.

Substitute $mg\cos \theta$ for ${F_{\rm{N}}}$ in the equation $f = \mu {F_{\rm{N}}}$ and rearrange to solve for $\mu$.

$\begin{array}{l}\\f = \mu mg\cos \theta \\\\\mu = \frac{f}{{mg\cos \theta }}\\\end{array}$

Substitute $5.3{\rm{ N}}$ for $f$, $2.30{\rm{ kg}}$ for $m$, $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$, and $36^\circ$ for $\theta$ in the equation $\mu = \frac{f}{{mg\cos \theta }}$ and calculate $\mu$.

$\begin{array}{c}\\\mu = \frac{{5.3{\rm{ N}}}}{{\left( {2.30{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\cos 36^\circ }}\\\\ = 0.29\\\end{array}$

Ans:

The magnitude of acceleration of the center of mass of the spherical shell is $3.46{\rm{ m/}}{{\rm{s}}^2}$.