**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{\

Given, The initial and final velocity with angular displacement, So, Angular acceleration is given…

Answered: The bit of a high-speed drill…

Similar questions

Starting from rest at t = 0, a wheel undergoes a constant angular acceleration. When t = 2.0 s, the angular velocity of the wheel is 5.0 rad/s.The acceleration continues until t = 20 s, when it abruptly ceases. Through what angle does the wheel rotate in the interval t= 0 to t= 40 s?
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.42 x 104 rad/s to an angular speed of 3.60 x 104 rad/s. In the process, the bit turns through 2.50 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 9.41 x 104 rad/s, starting from rest?
A 40.0-cm diameter disk rotates with a constant angular acceleration of 3.00 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time. (a) At t = 2.50 s, find the angular speed of the wheel. rad/s (b) At t = 2.50 s, find the magnitude of the linear velocity and tangential acceleration of P. linear velocity = m/s tangential acceleration = m/s2 (c) At t = 2.50 s, find the position of P (in degrees, with respect to the positive x-axis). ° counterclockwise from the +x-axis
Starting from rest at t = 0, a wheel undergoes a constant acceleration from t = 0 to t = 10 s. When t = 3.0 s, the angular velocity of the wheel is 6.0 rad/s. Through what angle does the wheel rotate from t = 0 to t = 20 s.
The angular acceleration of a wheel is a = 8.0t4 - 6.0t?, with a in radians per second-squared and t in seconds. At time t = 0, the wheel has an angular velocity of + 4 rad/s and an angular position of + 6 rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s). (a) Number i t5- i t3+ i Units (b) Number i t6- i t4+ i t+ Units
The angular acceleration of a wheel is α = 8.0t4 – 7.0t2, with αα in radians per second-squared and t in seconds. At time t = 0, the wheel has an angular velocity of +8 rad/s and an angular position of +1 rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).
A dentist's drill starts from rest. After 4.30 s of constant angular acceleration it turns at a rate of 2.05 ✕ 104 rev/min. (a) Find the drill's angular acceleration. rad/s2(b) Determine the angle (in radians) through which the drill rotates during this period.
The angular acceleration of a wheel is a = 9.0t4 - 4.0t2, with a in radians per second-squared and t in seconds. At time t = 0, the wheel has an angular velocity of +6 rad/s and an angular position of +8 rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s). (a) Number i t5- i t3+ Units (b) Number i t6. i t4+ t+ i Units
D V 24,919 :0 F2 X 3 3. A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.45 x 104 rad/s to an angular speed of 3.75 x 104 rad/s. In the process, the bit turns through 2.26 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.10 x 10¹ rad/s, starting from rest? X 2.135 s E D 20 F3 C 4. A 115 kg scaffold is 6.60 m long. It is hanging with two wires, one from each end. A 600 kg box sits 1.70 m from the SOIT P by m 04 M $ 4 7 R 208 F4 F V % 5 T F5 G ^ 6 B tv ♫ e MacBook Air F6 Y H & 7 N AA F7 J * 00 8 M DII FB - ( 9 K V DD F9 O F10 P DE zoom A C F11 EL = ? PI vaies or roe nuffer a 4) F12 delete re
A wheel rotates with a constant angular acceleration of 6.0 rad/s². During a certain time interval, its angular displacement is 6.0 rad. At the end of the interval, its angular velocity is 12.0 rad/s. Its angular velocity at the beginning of the interval is O 1.0 rad/s O 12 rad/s O 6.0 rad/s O O rad/s O 8.5 rad/s Save for Later Submit Answer
A playground merry-go-round has a radius of 3.5 m and a rotational inertia of 720 kgˑm2. It is initially spinning at 0.60 rad/s when a 25 kg child crawls from the center to the rim. When the child reaches the rim the angular velocity of the merry-go-round in rad/s is:
A figure skater performs a twisting jump. She rotates around her longitudinal axis 2 times in 0.65 s. What was her average angular velocity in radians/second? A 720rad/s B 19.3rad/s C 4? rad/s D 9.67 rad/s E 1107 rad/s

SEE MORE QUESTIONS