Chapter 9, Problem 90P

(a)

To determine

The acceleration of the center of mass of the spherical shell

Answer to Problem 90P

  $\frac{\text{3g}\text{sin}\text{θ}}{5}$

Explanation of Solution

Given:

The coefficient of static friction = ${\text{μ}}_{\text{s}}$

Angle of inclination = $\text{θ}$

Formula used:

Torque is defined as,

  $\text{τ}\text{=}\text{r}\text{F}\text{sin}\varphi$

F is the applied force on the object and r is the position vector from axis of rotation to the applied force

  $\varphi$ is the angle between r and F

Acceleration of the object in terms of angular acceleration and radius of rotation is defined as,

  $\text{a}\text{=}\text{Iα}$

Here I is the moment of inertia and $\text{α}$ is the angular acceleration.

Calculation:

Consider the static friction on the shell is ${\text{f}}_{\text{s}}$

Torque on the shell by static friction,

  $\text{τ}\text{=}\text{r}{\text{f}}_{\text{s}}$

Now, by second law of motion for rotation:

  $\text{τ}\text{=}\text{I}\text{α}$

  ${\text{f}}_{\text{s}}\text{r}\text{=}\left(\frac{\text{2}}{\text{3}}{\text{mr}}^{\text{2}}\right)\left(\frac{\text{a}}{\text{r}}\right)$

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}}{\text{3}}\text{ma}\mathrm{...}\text{(1)}$

Net force along the x axis,

  $\Sigma {\text{F}}_{\text{x}}\text{=}\text{mg}\text{sinθ}\text{-}{\text{f}}_{\text{s}}$

By second law of motion

  $\Sigma {\text{F}}_{\text{x}}\text{=}\text{m}\text{a}$

  $\text{mg}\text{sinθ}\text{-}{\text{f}}_{\text{s}}\text{=}\text{m}\text{a}$

  $\text{mg}\text{sinθ}\text{-}\frac{\text{2}}{\text{3}}\text{ma}\text{=}\text{m}\text{a}$

  $\frac{\text{2}}{\text{3}}\text{ma}+\text{m}\text{a}=\text{mg}\text{sinθ}$

  $\frac{5}{\text{3}}\text{ma}=\text{mg}\text{sinθ}$

  $\text{a}=\frac{3\text{g}\text{sinθ}}{5}$

Conclusion:

The center of mass of the spherical shell is $\frac{3\text{g}\text{sinθ}}{5}$ .

(b)

To determine

The frictional force acting on the ball

Answer to Problem 90P

The frictional force acting on the ball is $\frac{\text{2}\text{m}\text{g}\text{sinθ}}{5}$ .

Explanation of Solution

Given:

From part a),

Expression for the frictional force,

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}}{\text{3}}\text{ma}$

Acceleration of the center of mass of the spherical shell,

  $\text{a}=\frac{3\text{g}\text{sinθ}}{5}$

Calculation:

Since expression of the static friction,

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}}{\text{3}}\text{ma}$

Substitute the values:

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}}{\text{3}}\text{m}\left(\frac{3\text{g}\text{sinθ}}{5}\right)$

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}\text{mg}\text{sinθ}}{5}$

Conclusion:

The static friction on the shell is $\frac{\text{2}\text{mg}\text{sinθ}}{5}$ .

(c)

To determine

The maximum angle of the inclination for which the ball rolls without slipping

Answer to Problem 90P

  ${\text{θ}}_{\text{max}}\text{=}{\text{tan}}^{\text{-1}}\left(\frac{{\text{5μ}}_{\text{s}}}{\text{2}}\right)$

Explanation of Solution

Given:

From part a),

Expression for the frictional force,

  ${\text{f}}_{\text{s}}\text{=}\frac{\text{2}}{\text{3}}\text{m}\text{a}$

Acceleration of the center of mass of the spherical shell,

  $\text{a}=\frac{3\text{g}\text{sinθ}}{5}$

Calculation:

Net force along the y axis,

  $\Sigma {\text{F}}_y=F_n-mg\cos \theta$

Since, there is no any acceleration along the y axis, so, ${\text{a}}_y=0$

  $\Sigma {\text{F}}_{\text{y}}\text{=}{\text{ma}}_{\text{y}}$

  ${\text{F}}_{\text{n}}\text{-}\text{mg}\text{cos}\text{θ}\text{=}\text{m}\text{(0)}$

  ${\text{F}}_{\text{n}}\text{=}\text{mg}\text{cos}\text{θ}$

Maximum static friction,

  ${\text{f}}_{\text{s,max}}\text{=}{\text{μ}}_{\text{s}}{\text{F}}_{\text{n}}$

  ${\text{f}}_{\text{s,}\text{max}}\text{=}{\text{μ}}_{\text{s}}\text{(mg}\text{cos}\text{θ)}$

Now, torque on the shell,

  ${\tau }_{\max }\text{=}({\text{μ}}_{\text{s}}\text{mg}\text{cos}\text{θ)r}$

By 2^nd law of motion for rotation, we get,

  ${\text{τ}}_{\text{max}}\text{=}\text{I}\text{α}$

  ${\text{(μ}}_{\text{s}}\text{mg}\text{cos}\text{θ)r}\text{=}\left(\frac{\text{2}}{\text{3}}{\text{mr}}^{\text{2}}\right)\left(\frac{{\text{a}}_{\max }}{\text{r}}\right)$

  ${\text{a}}_{\text{max}}\text{=}\frac{{\text{3(μ}}_{\text{s}}\text{g}\text{cos}\text{θ)}}{\text{2}}$

Now, net force along the x axis,

  $\Sigma {\text{F}}_{\text{x}}\text{=}\text{mg}\text{sin}\text{θ}\text{-}{\text{f}}_{\text{s,max}}$

Now, by 2^nd law of motion, we get,

  $\text{mg}\text{sin}\text{θ}\text{-}{\text{f}}_{\text{s,max}}\text{=}\text{m}{\text{a}}_{\max }$

Now, let’s plug the value of $f_{s,\max }$ and a in the above equation, we get,

  $\text{mg}\text{sin}\text{θ}\text{-}{\text{μ}}_{\text{s}}\text{(mg}\text{cos}\text{θ)}\text{=}\text{m}\left(\frac{{\text{3(μ}}_{\text{s}}\text{g}\text{cos}\text{θ)}}{\text{2}}\right)$

  $\text{sin}\text{θ}={\text{μ}}_{\text{s}}\text{cos}\text{θ}+\frac{{\text{3(μ}}_{\text{s}}\text{cos}\text{θ)}}{2}$

  $\text{sin}\text{θ}=\frac{{\text{5μ}}_{\text{s}}\text{cos}\text{θ}}{2}$

  $\frac{\text{sin}\text{θ}}{\cos \text{θ}}=\frac{{\text{5μ}}_{\text{s}}}{2}$

  $\tan \theta =\frac{{\text{5μ}}_{\text{s}}}{2}$

  $\theta ={\tan }^{-1}\left(\frac{{\text{5μ}}_{\text{s}}}{2}\right)$

At the maximum acceleration, $\theta$ would be its maximum value. So, maximum angle:

  ${\theta }_{\max }={\tan }^{-1}\left(\frac{{\text{5μ}}_{\text{s}}}{2}\right)$

Conclusion:

The maximum angle is ${\tan }^{-1}\left(\frac{{\text{5μ}}_{\text{s}}}{2}\right)$ .