Chapter 10, Problem 16A

Biker Bob rides his bicycle inside the rotating space station at the speeds and directions given. The tangential speed of the floor of the station is 10 m/s clockwise.

a. Rank his speeds from highest to lowest relative to the stars.

b. Rank the normal forces on Bob from largest to smallest.

To determine

The rank of the speed of the biker from highest to lowest relative to the stars.

Answer to Problem 16A

The rank of the speed of the biker from highest to lowest relative to the stars is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Explanation of Solution

Given info:

The tangential speed of the floor of the station is $10\text{m}/\text{s}$ .

Formula used:

The expression for the speed for the objects moving in same direction as follows:

  $v=v_{\text{1}}-v_{\text{2}}$

Here, $v_{\text{1}}$ is the speed of the station and $v_{\text{2}}$ is the speed of the object.

Calculation:

The speed of the biker for case A as follows:

  $\begin{array}{l}v=v_{\text{1}}-v_{\text{2}}\\ v_{\text{A}}=\left(10\text{m}/\text{s}\right)-\left(6\text{m}/\text{s}\right)\\ v_{\text{A}}=4\text{m}/\text{s}\end{array}$

The speed of the biker for case B as follows:

  $\begin{array}{l}v=v_{\text{2}}-v_{\text{1}}\\ v_{\text{B}}=\left(10\text{m}/\text{s}\right)-\left(4\text{m}/\text{s}\right)\\ v_{\text{B}}=6\text{m}/\text{s}\end{array}$

The speed of the biker for case C as follows:

  $\begin{array}{l}v=v_{\text{1}}-v_{\text{2}}\\ v_{\text{C}}=\left(10\text{m}/\text{s}\right)-\left(-4\text{m}/\text{s}\right)\\ v_{\text{C}}=14\text{m}/\text{s}\end{array}$

By comparing the values, the speed of the biker is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Hence, the rank of the speed of the biker from highest to lowest relative to the stars is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Conclusion:

Thus, the rank of the speed of the biker from highest to lowest relative to the stars is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

To determine

The rank of the normal forces on Bob from largest to smallest.

Answer to Problem 16A

The rank of the normal forces on Bob from largest to smallest is $N_{\text{C}}>N_{\text{B}}>N_{\text{A}}$ .

Explanation of Solution

Given info:

The tangential speed of the floor of the station is $10\text{m}/\text{s}$ .

Formula used:

The expression for the normal force on the bicycle as follows:

  $N=\frac{m{v}^2}{r}$

Here, $m$ is the mass of the rider, $v$ is the speed of the rider relative to the stars, and $r$ is the radius.

Calculation:

Refer part (a).

The speed of the biker for case A $v_{\text{A}}=4\text{m}/\text{s}$ .

The speed of the biker for case B $v_{\text{B}}=6\text{m}/\text{s}$ .

The speed of the biker for case C $v_{\text{C}}=14\text{m}/\text{s}$ .

The normal force for case A as follows:

  $\begin{array}{l}N=\frac{m{v}^2}{r}\\ N_{\text{A}}=\frac{m{\left(4\text{m}/\text{s}\right)}^2}{r}\\ N_{\text{A}}=\frac{16m}{r}{\left(\text{m}/\text{s}\right)}^2\end{array}$

The normal force for case B as follows:

  $\begin{array}{l}N=\frac{m{v}^2}{r}\\ N_{\text{B}}=\frac{m{\left(6\text{m}/\text{s}\right)}^2}{r}\\ N_{\text{B}}=\frac{36m}{r}{\left(\text{m}/\text{s}\right)}^2\end{array}$

The normal force for case C as follows:

  $\begin{array}{l}N=\frac{m{v}^2}{r}\\ N_{\text{C}}=\frac{m{\left(14\text{m}/\text{s}\right)}^2}{r}\\ N_{\text{C}}=\frac{196m}{r}{\left(\text{m}/\text{s}\right)}^2\end{array}$

By comparing the values, the normal force is $N_{\text{C}}>N_{\text{B}}>N_{\text{A}}$ .

Conclusion:

Thus, the rank of the normal forces on Bob from largest to smallest is $N_{\text{C}}>N_{\text{B}}>N_{\text{A}}$ .