The bit of a high-speed drill accelerates from an angular speed of 2.35 x 102 rad/s to an angular speed of 5.65 x 102 rad/s. In the process, the bit turns through 275.5 Rev. assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 9.88 x 102 rad/s, starting from 1.05 x 102 rad/s?
The bit of a high-speed drill accelerates from an angular speed of 2.35 x 102 rad/s to an angular speed of 5.65 x 102 rad/s. In the process, the bit turns through 275.5 Rev. assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 9.88 x 102 rad/s, starting from 1.05 x 102 rad/s?
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![**Problem Statement:**
The bit of a high-speed drill accelerates from an angular speed of \( 2.35 \times 10^{2} \) rad/s to an angular speed of \( 5.65 \times 10^{2} \) rad/s. In the process, the bit turns through 275.5 revolutions. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of \( 9.88 \times 10^{2} \) rad/s, starting from \( 1.05 \times 10^{2} \) rad/s?
**Explanation:**
To solve such a problem, we will use the principles of rotational kinematics. Here are the steps to consider:
1. **Identify the Given Variables:**
- Initial angular speed (\( \omega_i \)) = \( 2.35 \times 10^{2} \) rad/s
- Final angular speed (\( \omega_f \)) = \( 5.65 \times 10^{2} \) rad/s
- Angular displacement (\( \theta \)) = 275.5 revolutions
- Convert revolutions to radians: \( \theta = 275.5 \times 2\pi \) radians.
2. **Key Equations:**
- Use the kinematic equation for angular motion:
\[
\omega_f^2 = \omega_i^2 + 2\alpha\theta
\]
- Where \( \alpha \) is the angular acceleration.
3. **Calculate Angular Acceleration:**
- Rearrange the equation to solve for \( \alpha \):
\[
\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}
\]
4. **Find the Time Taken to Reach Maximum Speed:**
- Maximum speed (\( \omega_{max} \)) = \( 9.88 \times 10^{2} \) rad/s
- New initial angular speed (\( \omega'_i \)) = \( 1.05 \times 10^{2} \) rad/s
- Using the equation:
\[
\omega_f = \omega'_i + \alpha t
\]
- Rearrange to solve for time (\( t \)):
\[
t = \frac{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae88911f-9dac-42fc-844f-67ccf125de46%2Fb5925429-232e-4bc7-9ed7-1a8eb83203f8%2Ftz73bl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The bit of a high-speed drill accelerates from an angular speed of \( 2.35 \times 10^{2} \) rad/s to an angular speed of \( 5.65 \times 10^{2} \) rad/s. In the process, the bit turns through 275.5 revolutions. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of \( 9.88 \times 10^{2} \) rad/s, starting from \( 1.05 \times 10^{2} \) rad/s?
**Explanation:**
To solve such a problem, we will use the principles of rotational kinematics. Here are the steps to consider:
1. **Identify the Given Variables:**
- Initial angular speed (\( \omega_i \)) = \( 2.35 \times 10^{2} \) rad/s
- Final angular speed (\( \omega_f \)) = \( 5.65 \times 10^{2} \) rad/s
- Angular displacement (\( \theta \)) = 275.5 revolutions
- Convert revolutions to radians: \( \theta = 275.5 \times 2\pi \) radians.
2. **Key Equations:**
- Use the kinematic equation for angular motion:
\[
\omega_f^2 = \omega_i^2 + 2\alpha\theta
\]
- Where \( \alpha \) is the angular acceleration.
3. **Calculate Angular Acceleration:**
- Rearrange the equation to solve for \( \alpha \):
\[
\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}
\]
4. **Find the Time Taken to Reach Maximum Speed:**
- Maximum speed (\( \omega_{max} \)) = \( 9.88 \times 10^{2} \) rad/s
- New initial angular speed (\( \omega'_i \)) = \( 1.05 \times 10^{2} \) rad/s
- Using the equation:
\[
\omega_f = \omega'_i + \alpha t
\]
- Rearrange to solve for time (\( t \)):
\[
t = \frac{\
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