1.25 cm. 65 ms. (a) What is the magnitude of the average angular velocity of the
ID: 1464531 • Letter: 1
Question
1.25 cm.
65 ms.
(a) What is the magnitude of the average angular velocity of the eye? (b) Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, then the rotation slows with a constant angular acceleration during the second half until it comes to rest. What is the magnitude of the angular acceleration of the eye? (c) What tangential acceleration would the contact-lens accelerometers record in this case?
I got that for part a the average angular velocity is 8 rad/s and that is the right answer. I also know that this question has been answered on chegg before, but the way that it is being solved is not getting me the correct answer for part b. It would be greatly appreciated if someone could walk me through part b.
Thank you.
A study was done observing the ability of the eye to rapidly rotate in order to follow a moving object by placing contact lenses that contain accelerometers on a subject's eye. The eyeball has radius1.25 cm.
Suppose that, while the subject watches a moving object, the eyeball rotates through 30.0° in a time interval of65 ms.
(a) What is the magnitude of the average angular velocity of the eye? (b) Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, then the rotation slows with a constant angular acceleration during the second half until it comes to rest. What is the magnitude of the angular acceleration of the eye? (c) What tangential acceleration would the contact-lens accelerometers record in this case?
I got that for part a the average angular velocity is 8 rad/s and that is the right answer. I also know that this question has been answered on chegg before, but the way that it is being solved is not getting me the correct answer for part b. It would be greatly appreciated if someone could walk me through part b.
Thank you.
Explanation / Answer
a) angular velocity w = theta/t
theta = 30 degrees = pi/6 radians
w = pi/6 / 65 ms = 8.05 rad/s
b) w0 =0
theta = 30/2 = 15 degrees = pi/12 radians ( for half)
t = 65/2 ms
theta = w0 t + 1/2 alpha t2
alpha = 2 theta/t2 = 2(pi/12) / (65/2 x 10-3)2
alpha = 495.46 rad/s2
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