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

Question

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

What is the tangential acceleration of the reference dot at t = 0.25 s?

What is the centripetal acceleration of the reference dot at t2 = 0.25 s?

What is the angular position of the reference dot at t3 = 1.0 s?

What is the speed of the reference dot at t4 = 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 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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