Chapter 17.2, Problem 17.70P

To determine

(a)

The velocity of the wheel’s center at time t.

Answer to Problem 17.70P

Velocity of wheel center at time t will be,

$V=\frac{r^{2}gtsin\beta }{r^2 +\overline{k^2}}<\beta$

Explanation of Solution

Given:

Wheel is initially at rest and it is relased from an indined surface from rest.

Wheel radius is $r$

The radius of gyration of the wheel is $\overline{k}$

Initial velocity $v_0=0$

Concept used:

Impulse momentum principle,

Moment of inertia

Assume rolling of the wheel without sliding

According to theimpulse-momentum principle,

Taking moment about the point of contact C,

$0+\left(mgt\right)rsin\beta =m\overline{v}r+I\omega$ ------------------(1)

Considering the rolling motion of the wheel,

$\begin{array}{l}\overline{v}=v_G=r\omega \hfill \\ \omega =\frac{\overline{v}}{r}\hfill \end{array}$

also a moment of inertia, $I=m\overline{k^2}$

putting values in equation (1),

$\begin{array}{l}\begin{array}{l}mgtrsin\beta =m\overline{v}r+I\omega \hfill \\ =m\overline{v}r+\frac{\overline{k^2}\overline{v}}{r}\hfill \\ =\overline{v}\left(mr+\frac{m\overline{k^2}}{r}\right)\hfill \\ mgtrsin\beta =\overline{v}\left(mr+\frac{m\overline{k^2}}{r}\right)\hfill \\ {\frac{mgtrsin\beta r}{mr+m\overline{k^2}}}^2=\overline{v}\hfill \end{array}\\ \overline{v}=\frac{r^{2}gtsin\beta }{r^2+\overline{k^2}}\end{array}$

Conclusion:

In this way we can calculate the velocity of the wheel at time t by the impulse-momentum principle and simple calculation,

$\overline{v}=\frac{r^{2}gtsin\beta }{r^2+\overline{k^2}}$

To determine

(b)

The coefficient of static friction needed to overcome the slipping of the wheel.

Answer to Problem 17.70P

Coefficient of friction between wheel and surface needed to overcome slipping of the wheel is

${\mu }_s=\frac{k^{2}tan\beta }{r^2+\overline{k^2}}$

Explanation of Solution

Given:

Wheel radius is $r$

The radius of gyration of the wheel is $\overline{k}$

Initial velocity $v_0=0$

Concept used:

Impulse momentum principle,

Moment of inertia

Assume rolling of the wheel without sliding

Parallel component of inclination

$\begin{array}{l}0+mgtsin\beta -ft=m\overline{v}\hfill \\ Ft=mgtsin\beta -m\left[\frac{r^{2}gtsin\beta }{r^2+\overline{k^2}}\right]\hfill \\ =\frac{mgtsin\beta }{\left(\frac{1–r^2}{r^2+\overline{k^2}}\right)}\hfill \\ Ft=\frac{\overline{k^2}}{r^2+\overline{k^2}mgtsin\beta }\hfill \end{array}$

The normal component of indignation

$\begin{array}{l}0+Nt–mgtcos\beta =0\hfill \\ Nt=mgtcos\beta \hfill \end{array}$

As we know thsat, the coefficient of friction is the ratio of force to the normal reaction between the two objects.

Thus we have,

$\begin{array}{l}Ft=\frac{\overline{k^2}}{r^2+\overline{k^2}mgtsin\beta }\hfill \\ Nt=mgtcos\beta \hfill \\ {\mu }_k=\frac{Ft}{Nt}=\frac{ k^{2}mgtsin\beta }{r^2+\overline{k^2}mgtcos\beta }\hfill \end{array}$

${\mu }_s=\frac{k^{2}tan\beta }{r^2+\overline{k^2}}$

Conclusion:

Hence, by calculating the normal and parallel component of indignation, we get the value of the coefficient of friction between wheel and slope is ${\mu }_s=\frac{k^{2}tan\beta }{r^2+\overline{k^2}}.$