During a certain time interval, the angular position of a swinging door is descr

Question

During a certain time interval, the angular position of a swinging door is described by theta = 4.99 + 9.4t + 1.91 t^2, where theta is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times. t = 0 theta = omega = alpha = t = 2.93 s theta = omega = alpha = An electric motor rotating a workshop grinding wheel at 1.16 times 10^2 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 1.94 rad/s^2. How long does it take the grinding wheel to stop? Note that the time required for the wheel to stop is the same as the time for the wheel to start from rest and reach the given angular speed with the same magnitude angular acceleration. s Through how many radians has the wheel turned during the time interval found in part (a)? rad

Explanation / Answer

given

= 4.99 + 9.4t + 1.91t2

t= 0

= 4.99

w= d/dt = 9.4 + 3.82 t

t= 0

w = 9.4 rad/s

dw/dt = = 3.82 rad/s^2

when t = 2.93 s

= 4.99 + 9.4(2.93) + 1.91(2.93)2 = 48.92 rad

w = d/dt = 9.4 + 3.82 t = 9.4 + 3.82 ( 2.93) = 20.59 rad/s

dw/dt = = 3.82 rad/s^2

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(2)

w = 116 ( 2 pi/60) = 12.14 rad/s

t = w2- w1/ alpha =0- 12.14/-1.94 = 6.25 s

= 1t + (1/2)t^2

   = 12.14(6.25) - 1/2 ( 1.94) ( 6.25)^2 = 37.88 rad

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