Chapter 8, Problem 129P

(a)

To determine

The magnitude of static frictional force exerted by road on each tire.

Answer to Problem 129P

The magnitude of static frictional force exerted by road on each tire is $\underset{\_}{f_{\text{road}}=62\text{N}\text{backwards}}$.

Explanation of Solution

Consider Newton’s second law to find the force exerted by road on bicycle, since it is the only external force acting on bicycle.

Write the Mathematical expression for Newton’s second law

${\displaystyle \sum F_x}=M{a}_x$ (I)

Here, $F_x$ is the force, $M$ is the mass, and $a_x$ is the acceleration

Both wheels of bicycle experience external force between road and wheel. Hence the expression for that force in terms of velocity is

$\begin{array}{c}2f_{\text{road}}=M\frac{\Delta v_x}{\Delta t}\\ =M\frac{\left(v_f-v_i\right)}{\Delta t}\end{array}$ (II)

Here, $\Delta t$ is the change in time, $v_f$ is the final velocity, and $v_i$ is the initial velocity

Therefore, the expression of frictional force is

$f_{\text{road}}=\frac{M\left(v_f-v_i\right)}{2\Delta t}$ (III)

Conclusion:

Substitute, $74\text{kg}$ for $M$, $0\text{m/s}$ for $v_i$, $75\text{m/s}$ for $v_f$, and $4.5\text{s}$ for $\Delta t$ in equation (III)

$\begin{array}{c}{f}_{\text{road}}=\frac{74\text{kg}\left(0-7.5\text{m/s}\right)}{2\left(4.5\text{s}\right)}\\ =-61.7\text{N}\\ \approx \text{62}\text{N,}\text{backwards}\end{array}$

Therefore, the magnitude of static frictional force exerted by road on each tire is $\underset{\_}{f_{\text{road}}=62\text{N}\text{backwards}}$.

(b)

To determine

The angular acceleration of two wheels.

Answer to Problem 129P

The angular acceleration of two wheels is $\underset{\_}{\alpha =4.8{\text{rad/s}}^2}$.

Explanation of Solution

Write the expression for angular speed

$\omega =\frac{v}{r}$ (IV)

Here, $\omega$ is the angular speed, $v$ is the linear velocity, $r$ is the radius

Write the expression for angular acceleration

$\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}$ (V)

Here, $\alpha$ is the angular acceleration, $\Delta t$ is the change in time, ${\omega }_f$ is the final angular speed, ${\omega }_i$ is the initial angular speed

Conclusion:

Substitute, $7.5\text{m/s}$ for $v$, $0.35\text{m}$ for $r$ in equation (VI)

$\begin{array}{c}\omega =\frac{7.5\text{m/s}}{0.35\text{m}}\\ =21.4\text{rad/s}\end{array}$

Substitute $0$ for ${\omega }_i$, $21.4\text{rad/s}$ for ${\omega }_f$, $4.5\text{s}$ for $\Delta t$ in equation (V)

$\begin{array}{c}\alpha =\frac{0-21.4\text{rad/s}}{4.5\text{s}}\\ =-4.75{\text{rad/s}}^2\\ \approx 4.8{\text{rad/s}}^2\end{array}$

Therefore, the angular acceleration of two wheels is $\underset{\_}{\alpha =4.8{\text{rad/s}}^2}$, opposite to the direction of original rotation.

(c)

To determine

The net torque acting on wheel.

Answer to Problem 129P

The net torque acting on wheel is $\underset{\_}{{\displaystyle \sum \tau }=0.76\text{N}\cdot \text{m}}$.

Explanation of Solution

Write the expression for net torque

${\displaystyle \sum \tau }=I\alpha$ (VI)

Here, $\tau$ is the torque, $I$ is the moment of inertia, $\alpha$ is the angular acceleration

Write the expression for moment of inertia

$I=m{r}^2$ (VII)

Here, $m$ is the mass, $r$ is the radius

Substitute equation (VII) in (VI)

${\displaystyle \sum \tau }=m{r}^2\alpha$ (VIII)

Conclusion:

Substitute, $1.3\text{kg}$ for $m$, $0.35\text{m}$ for $r$, $-4.76{\text{rad/s}}^2$ for $\alpha$ in equation (VIII)

$\begin{array}{c}{\displaystyle \sum \tau }=\left(1.3\text{kg}\right){\left(0.35\text{m}\right)}^2\left(-4.76{\text{rad/s}}^2\right)\\ =-0.758\text{N}\cdot \text{m}\\ \approx \text{0}\text{.76}\text{N}\cdot \text{m}\text{opposite to the direction of original rotation}\end{array}$

Therefore, the net torque acting on wheel is $\underset{\_}{{\displaystyle \sum \tau }=0.76\text{N}\cdot \text{m}}$, opposite to the direction of original rotation.

(d)

To determine

The normal force applied to a wheel by each of the brake pads.

Answer to Problem 129P

The normal force applied to a wheel by each of the brake pads is $\underset{\_}{f_{\text{brake}}=35\text{N}}$.

Explanation of Solution

There are two components of net torque acting on each wheel, one is static friction force acting between road and wheel, and the other is the kinetic friction force acting on brake pads.

The direction of both forces will be opposite to each other. If the bicycle turns to right, static friction force will be in clockwise direction, and kinetic friction will be in counter clockwise direction.

Write the expression for net force in terms components

${\displaystyle \sum \tau }=-f_{\text{road}}r+2f_{\text{brake}}r$ (IX)

Rearrange equation (IX) to obtain expression for $f_{\text{brake}}$

$f_{\text{brake}}=\frac{{\displaystyle \sum \tau }+f_{\text{road}}r}{2r}$ (X)

Write the expression for normal force

$N=\frac{f_{\text{brake}}}{\mu }$ (XI)

Conclusion:

Substitute, $0.758\text{N}\cdot \text{m}$ for ${\displaystyle \sum \tau }$, $0.35\text{m}$ for $r$, and $61.7\text{N}$ for  in equation (X)

$\begin{array}{c}{f}_{\text{brake}}=\frac{0.758\text{N}\cdot \text{m}+\left(61.7\text{N}\right)\left(0.35\text{m}\right)}{2\left(0.35\text{m}\right)}\\ =31.9\text{N}\end{array}$

Substitute $0.9$ for $\mu$, $31.9\text{N}$ for $f_{\text{brake}}$ in equation (XI)

$\begin{array}{c}N=\frac{31.9\text{N}}{0.9}\\ =35\text{N}\end{array}$

Therefore, the normal force applied to a wheel by each of the brake pads is $\underset{\_}{f_{\text{brake}}=35\text{N}}$.