When a carpenter shuts off his circular saw, the 10.0-inchdiameter blade slows f

Question

When a carpenter shuts off his circular saw, the 10.0-inchdiameter blade slows from 4550 rpm to zero in3.00 s. What is the angular acceleration of theblade? What is the distance traveled by a point onthe rim of the blade during the deceleration? What is the magnitude of the net displacement of a point onthe rim of the blade during the deceleration? When a carpenter shuts off his circular saw, the 10.0-inchdiameter blade slows from 4550 rpm to zero in3.00 s. What is the angular acceleration of theblade? What is the angular acceleration of theblade? What is the distance traveled by a point onthe rim of the blade during the deceleration? What is the distance traveled by a point onthe rim of the blade during the deceleration? What is the magnitude of the net displacement of a point onthe rim of the blade during the deceleration?

Explanation / Answer

Given that the diameter of the blade is d = 10 inch = 0.254m Initial angular velocity is 1 = 4550 rpm = 476.23rad/s Final angular velocity is 2 = 0 Time taken is t = 3.0s ------------------------------------------------------ The angular acceleration is = ( 2 - 1) / t                                          = --------- rad/s2 The angular displacement is = (22 - 12 ) /2                                            = ------------ rad Then the distance traveled by the point on the rim is x =(d/2)*                                                                              = ----------- m Number of revolutions is N = x / 2r = x / 2(d/2)= x / d =---------

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