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:2. A how far in degrees did the hand travel during the five rotations?B. How far in radians did the hand travel during the five rotations?C. How far in meters did the hand travel during the five rotations?3. A. What was the average angular speed (degrees/s and rad/s) of the hand?B. What was the average linear speed (m/s) of the hand?C. Are the answers to a and b the same or different? Explain.4. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the hand. How do you know?B. What was the average centripetal acceleration (m/s squared) of the hand?C. Are the answers to a and b the same or different. Explain.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

Given:

Time taken for 5 rotations = 5.15 seconds

Time for 1 rotation = 1.07 seconds

Distance from shoulder to elbow = 29 cm

Distance from shoulder to the middle of the hand = 57 cm

Let's use the information above to answer the following questions.

Question 2:

Let's determine how far in degrees the hand travelled during the five rotations.

In one full rotation, we have 360 degrees.

Thus, 5 full rotations = 5 * 360 = 1800 degrees

Therefore, in 5 full rotations, the hand travelled 1800 degrees.

B. In radians, we have:

180 degrees = π rad

`1800\degree=(\pi)/(180)\ast1800=10\pi\text{ radians}`

C. To find the distance in meters, we have:

Distance from elbow to shoulder = 29 cm = 0.29 meters

`2\pi\ast5\ast0.29=9.11\text{meters}`

Therefore, the hand travelled 9.11 meters during the five rotations.

Question 3:

A. To find the average angular speed, apply the formula:

`\begin{gathered} w=(10\pi)/(t)\text{ (rad/s)} \\ \\ w=(1800)/(t)\text{ (degre}es\text{/s)} \end{gathered}`

Where t = 5.15 seconds

Thus, we have:

`\begin{gathered} w=(10\pi)/(5.15)=6.1\text{ rad/s} \\ \\ w=(1800)/(5.15)=349.5\text{ degre}es\text{/s} \end{gathered}`

B. Average linear speed of the hand.

To find the average linear speed of the hand, we have:

`v=(10\pi r)/(t)=(10\pi)/(5.15)\ast(1)/(2)=3.05\text{ m/s}`

C. The average angular speed and average linear speed are the same