Required information The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has radius 1.25 cm, watches a moving object her eyeball rotates through 20.0 in a time interval of 71.3 ms. Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, and 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? 68 6115 radis2
Required information The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has radius 1.25 cm, watches a moving object her eyeball rotates through 20.0 in a time interval of 71.3 ms. Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, and 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? 68 6115 radis2
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![### Understanding Angular Acceleration of the Eye
#### Required Information
The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has a radius of 1.25 cm, watches a moving object, her eyeball rotates through 200° in a time interval of 713 ms.
#### Task
Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, and 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?
#### Calculation
To solve this, consider the given information:
- Eyeball radius: 1.25 cm
- Total rotation: 200°
- Time interval: 713 ms
- The motion is split into two equal halves: acceleration and deceleration.
The angular acceleration (α) can be calculated considering that the eye goes through a uniform acceleration phase for the first half of the time and decelerates uniformly during the second half of the time.
\[
\theta = 200° \quad (\text{Angular displacement})
\]
First, convert the angular displacement to radians:
\[
200° \times \frac{\pi \text{ rad}}{180°} = \frac{200\pi}{180} \approx 3.49 \text{ rad}
\]
The total time given is 713 ms, which is:
\[
713 \text{ ms} = 0.713 \text{ s}
\]
Therefore, half the time interval (for acceleration or deceleration phases) is:
\[
t = \frac{0.713}{2} \text{ s} = 0.3565 \text{ s}
\]
Using the kinematic equation for angular motion:
\[
\theta = \frac{1}{2} \alpha t^2
\]
\(\alpha\) can be isolated and solved:
\[
\alpha = \frac{2 \theta}{t^2}
\]
Substitute \(\theta = 3.49 \text{ rad}\) and \(t = 0.3565 \text{ s}\):
\[
\alpha = \frac{2 \times 3.49 \text{ rad}}{(0.3565 \text{ s})^2}
\]
\[
\alpha =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff7f72a6-27ca-4fc6-a195-84c52d046220%2Fe6c9c0a1-0bc8-4634-acbf-0da031ae6a24%2Fdygrqf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Angular Acceleration of the Eye
#### Required Information
The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has a radius of 1.25 cm, watches a moving object, her eyeball rotates through 200° in a time interval of 713 ms.
#### Task
Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, and 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?
#### Calculation
To solve this, consider the given information:
- Eyeball radius: 1.25 cm
- Total rotation: 200°
- Time interval: 713 ms
- The motion is split into two equal halves: acceleration and deceleration.
The angular acceleration (α) can be calculated considering that the eye goes through a uniform acceleration phase for the first half of the time and decelerates uniformly during the second half of the time.
\[
\theta = 200° \quad (\text{Angular displacement})
\]
First, convert the angular displacement to radians:
\[
200° \times \frac{\pi \text{ rad}}{180°} = \frac{200\pi}{180} \approx 3.49 \text{ rad}
\]
The total time given is 713 ms, which is:
\[
713 \text{ ms} = 0.713 \text{ s}
\]
Therefore, half the time interval (for acceleration or deceleration phases) is:
\[
t = \frac{0.713}{2} \text{ s} = 0.3565 \text{ s}
\]
Using the kinematic equation for angular motion:
\[
\theta = \frac{1}{2} \alpha t^2
\]
\(\alpha\) can be isolated and solved:
\[
\alpha = \frac{2 \theta}{t^2}
\]
Substitute \(\theta = 3.49 \text{ rad}\) and \(t = 0.3565 \text{ s}\):
\[
\alpha = \frac{2 \times 3.49 \text{ rad}}{(0.3565 \text{ s})^2}
\]
\[
\alpha =
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