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 2Ï€ (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)(2Ï€ 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
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