Chapter 12, Problem 21A

Perky rides at different radial distances from the center of a turntable that rotates at a fixed rate. His distances and tangential speeds at three different locations are as follows.

(A) $r=15cm,v=7.5cm/s$

(B) $r=10cm,v=7.5cm/s$

(C) $r=5.0cm,v=2.5cm/s$

a. Rank from greatest to least, Perky’s angular momenta.

b. Rank from greatest to least, the amounts of friction needed to keep Perky from sliding off.

To determine

The rank of the Person P’s angular momenta from greatest to least.

Answer to Problem 21A

The rank of the Person P’s angular momenta from greatest to least is $L_{\text{A}}>L_{\text{B}}>L_{\text{C}}$ .

Explanation of Solution

Formula used:

The expression for the angular momenta as follows:

  $L=mvr$

Here, $m$ is the mass, $v$ is the tangential speed, and $r$ is the radius of the curved path.

Calculation:

Find the angular momenta for case A as follows:

  $\begin{array}{l}{L}_{\text{A}}=m{v}_{\text{A}}{r}_{\text{A}}\\ L_{\text{A}}=m\left(7.5\text{cm}/\text{s}\right)\left(15\text{ cm}\right)\\ L_{\text{A}}=112.5m\left({\text{cm}}^2/\text{s}\right)\end{array}$

Find the angular momenta for case B as follows:

  $\begin{array}{l}{L}_{\text{B}}=m{v}_{\text{B}}{r}_{\text{B}}\\ L_{\text{B}}=m\left(5\text{cm}/\text{s}\right)\left(10\text{ cm}\right)\\ L_{\text{B}}=50m\left({\text{cm}}^2/\text{s}\right)\end{array}$

Find the angular momenta for case A as follows:

  $\begin{array}{l}{L}_{\text{C}}=m{v}_{\text{C}}{r}_{\text{C}}\\ L_{\text{C}}=m\left(2.5\text{cm}/\text{s}\right)\left(5\text{ cm}\right)\\ L_{\text{C}}=12.5m\left({\text{cm}}^2/\text{s}\right)\end{array}$

By comparing the values, the angular momenta is $L_{\text{A}}>L_{\text{B}}>L_{\text{C}}$ .

Conclusion:

Thus, the rank of Person P’s angular momenta from greatest to least is $L_{\text{A}}>L_{\text{B}}>L_{\text{C}}$ .

To determine

The rank of the amounts of friction needed to keep Person P’s from sliding off from greatest to least.

Answer to Problem 21A

The rank of the amounts of friction needed to keep Person P’s from sliding off from greatest to least is $f_{\text{A}}>f_{\text{B}}>f_{\text{C}}$ .

Explanation of Solution

Formula used:

The expression for the friction as follows:

  $f=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the tangential speed, and $r$ is the radius of the curved path.

Calculation:

Find the friction needed for case A as follows:

  $\begin{array}{l}{f}_{\text{A}}=\frac{m{v}_{\text{A}}^2}{r_{\text{A}}}\\ f_{\text{A}}=\frac{m{\left(7.5\text{cm}/\text{s}\right)}^2}{\left(15\text{ cm}\right)}\\ f_{\text{A}}=3.75m\text{ N}\end{array}$

Find the friction needed for case B as follows:

  $\begin{array}{l}{f}_{\text{B}}=\frac{m{v}_{\text{B}}^2}{r_{\text{B}}}\\ f_{\text{B}}=\frac{m{\left(5\text{cm}/\text{s}\right)}^2}{\left(10\text{ cm}\right)}\\ f_{\text{B}}=2.5m\text{ N}\end{array}$

Find the angular momenta for case A as follows:

  $\begin{array}{l}{f}_{\text{C}}=\frac{m{v}_{\text{C}}^2}{r_{\text{C}}}\\ f_{\text{C}}=\frac{m{\left(2.5\text{cm}/\text{s}\right)}^2}{\left(\text{5 cm}\right)}\\ f_{\text{C}}=1.25m\text{ N}\end{array}$

By comparing the values, the friction needed is $f_{\text{A}}>f_{\text{B}}>f_{\text{C}}$ .

Conclusion:

Thus, the rank of amounts of friction needed to keep Person P’s from sliding off from greatest to least is $f_{\text{A}}>f_{\text{B}}>f_{\text{C}}$ .