Chapter 5, Problem 14P

To determine

The force exerted to pull one another if it is assumed that the arms of the subjects are each $\text{0}\text{.80}\text{m}$ long and their individual masses are $\text{55}\text{.0}\text{kg}$ .

Answer to Problem 14P

Solution:

Skaters pulleach other byforce $\text{277}\text{.20}\text{N}$ .

Explanation of Solution

Given:

One revolution every $\text{2}\text{.5}\text{s}$

Mass of each skater = 55.0 kg

Arm length of each skater = $\text{0}\text{.80}\text{m}$

Formula Used:

Angular speed is defined as the angle of rotation per second.

Centripetal force, $\text{=}m.a_c=m{\omega }^{2}r$

Where, $\omega$ is the angular velocity.

Calculation:

  

Pull on each other provide necessary centripetal force to move in a circle.

  $\begin{array}{l}F=m{\omega }^{2}r\\ \omega =\frac{2\pi }{2.5}=2.51\text{ rad/s }\\ r=0.80\text{m }\end{array}$

r is the arm length.

  $\begin{array}{l}F=\left(\text{55}\text{.0}\right)\text{×}{\left(\text{2}\text{.51}\right)}^{\text{2}}\text{×}\left(\text{0}\text{.80}\right)\text{N}\\ \text{ =}\text{277}\text{.20}\text{N}\end{array}$

Conclusion:Hence, skaters pull each other by force $\text{277}\text{.20}\text{N}$ .