Question

Multiple part question Here are the needed details:Five rotations took 5.15 seconds 1 rotation took 1.07s Distance from shoulder to elbow is 29 cm distance from shoulder to middle of the hand is 57cm.Questions:5. A. How far (degrees and rad) did the elbow travel during the five rotations?B. How far (m) did the elbow travel during the five rotations?C. How do these compare to the hand? Why are they the same and or/different?6. A. What was the average angular speed (degrees/s and rad/s) of the elbow?B. What was the average linear speed (m/s) of the elbow?C. How do these compare to the hand? Why are they the same and or/different?7. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the elbow?B. What was the average centripetal acceleration (m/s squared) of the elbow?C. How do these compare to the hand? Why are they the same and or/ different?

Answer 1

5. A. In one rotation the arm will travel 2π.

Therefore in 5 rotations it will travel,

`5*2\pi=10\pi\text{ rad}`

Which is equivalent to,

`\begin{gathered} 10\pi=10*180^(\circ)^{} \\ =1800^(\circ) \end{gathered}`

Therefore, in five revolution elbow will travel 10π rad or 1800 degree.

B. The length of the elbow is given as,

`\begin{gathered} r=29\text{ cm } \\ =0.29\text{ m} \end{gathered}`

Therefore distance travelled by the elbow is given as,

`\begin{gathered} s=r\theta \\ =0.29\text{ m}*10\pi \\ =9.11\text{ m} \end{gathered}`

Therefore, elbow will travel a distance of 9.11 m.

C. The length of the hand is given as,

`\begin{gathered} L=2*57\text{ cm } \\ =114\text{ cm} \\ =1.14\text{ m} \end{gathered}`

Therefore distance travelled by the hand is given as,

`\begin{gathered} s^(\prime)=\theta L \\ =10\pi*1.14\text{ m} \\ =35.81\text{ m} \end{gathered}`

Therefore, hand will travell a distance of 35.81 m.