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: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

(6)

The formula for the average angular speed of the elbow is given as

`v_(avg)=(\theta_t)/(T_(total))`

Substitute the values in the above expression as

`\begin{gathered} v_(avg)=(5*360^0+1*360^0)/(5.15+1.07) \\ =347.266\text{ degre}e\text{/s} \\ =6.06\text{ rad/s} \end{gathered}`

(B)

The average linear speed of the elbow is calculated as

`\begin{gathered} v_s=v_(avg)* r \\ =6.06*0.29 \\ =1.75\text{ m/s} \end{gathered}`

(C)

The distance between the shoulder and the middle of the hand is 57 cm or approximately twice times. Hence, the average linear speed of the hand will be double of the elbow