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Answer :
Final answer:
A shaft of diameter 0.30 m is rotating at 2000 rev/min. Its angular velocity is 209.44 rad/s and the linear speed of a point on the circumference of the shaft is 31.42 m/s. The angular acceleration required to increase the shaft speed to 3000 rev/min in a period of seconds can be calculated using the formula 209.44 rad/s² / t, and the linear acceleration of a point on the circumference of the shaft is 31.42 m/s² / t.
Explanation:
A. Angular velocity: To convert from rev/min to radians per second, we need to multiply by 2π (the conversion factor from rev to radians) and divide by 60 (the conversion factor from minutes to seconds). So, the angular velocity is:
Angular velocity = (2000 rev/min)(2π rad/rev)(1 min/60 s) = 209.44 rad/s
B. Linear speed: The linear speed of a point on the circumference of the shaft is the product of the angular velocity and the radius of the shaft. So, the linear speed is:
Linear speed = (209.44 rad/s)(0.15 m) = 31.42 m/s
C. Angular acceleration: The change in angular velocity is the final angular velocity minus the initial angular velocity. So, the angular acceleration is:
Angular acceleration = (3000 rev/min - 2000 rev/min)(2π rad/rev)(1 min/60 s) / t = 209.44 rad/s² / t
D. Linear acceleration: The linear acceleration of a point on the circumference of the shaft is the product of the angular acceleration and the radius of the shaft. So, the linear acceleration is:
Linear acceleration = (209.44 rad/s² / t)(0.15 m) = 31.42 m/s² / t
Learn more about Angular velocity, linear speed, angular acceleration, linear acceleration here:
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