A computer disk is 8.0 cm in diameter. A reference dot on the edge of the disk i

Question

A computer disk is 8.0 cm in diameter. A reference dot on the edge of the disk is initially located at theta 45 degrees. The disk accelerates steadily for 0.50 s, reaching 2000 rpm, then coasts at steady angular velocity for another 0.50 s.

What is the tangential acceleration of the reference dot at t = 0.25 s?
What is the centripetal acceleration of the reference dot at t = 0.25 s?
What is the angular position of the reference dot at t = 1.0 s?
What is the speed of the reference dot at t = 1.0 s?

Explanation / Answer

Initial position of the dot is ?0 = 45degrees = 1.57 rad initial angular speed is 0 rad/s angular speed after 0.5s is 2000rpm = 209rad/s then the angular acceleration of the disc is (209rad/s - 0)/(0.5s) = 418rad/s^2 then the tangential acceleration of the reference dot is at = (418rad/s^2)(0.04m) = 16.72m/s^2 and the tangential speed of the reference dot at 0.25s is v = 0+ (16.72m/s^2)(0.25s) = 4.18m/s then the centripetal acceleration of the reference dot at 0.25s is ac = (4.18m/s)^2/(0.04m) =436.81m/s^2 angular position of the dot after 1s will be ? = 1.57rad + (1/2)(418rad/s^2)(0.5s)^2 + (209rad/s)(0.5s) = 158.32rad tangential speed of the reference dot at 1s is the speed at 0.5s given by v = 0+ (16.72m/s^2)(0.5s) = 8.36m/s

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