Respuesta :
Answer:
da = 66 rev
The tub turn 66 revolutions while it is in motion
Question:
The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 9.00 s, at which time it is turning at 6.00 rev/s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 13.0 s. Through how many revolutions does the tub turn while it is in motion?
Explanation:
Given:
Acceleration Time t1 = 9.00s
Deceleration Time t2 = 13.00s
Maximum angular velocity w2= 6.00rev/s
Initial and final angular speed w1= 0rev/s
Calculating the average angular velocity wa;
wa = (w1+w2)/2
wa = (6+0)/2
wa = 3 rev/s
Angular distance covered da= average angular velocity × time
da = wat1 + wat2
da = 3×9 + 3×13
da = 27+39
da = 66 rev
The tub turn 66 revolutions while it is in motion
Answer:
[tex]n_{tot} = 66\,rev[/tex]
Explanation:
The total of revolutions during the whole process is:
[tex]n_{tot} = n_{1} + n_{2}[/tex]
The acceleration of each stage is determined below:
Acceleration:
[tex]\ddot n_{1} = \frac{\dot n_{1}}{t_{1}}[/tex]
[tex]\ddot n_{1} = \frac{6\,\frac{rev}{s} }{9\,s}[/tex]
[tex]\ddot n_{1} = \frac{2}{3}\,\frac{rev}{s^{2}}[/tex]
Deceleration:
[tex]\ddot n_{2} = -\frac{\dot n_{1}}{t_{2}}[/tex]
[tex]\ddot n_{2} = -\frac{6\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{2} = - \frac{6}{13}\,\frac{rev}{s^{2}}[/tex]
The total of revolutions during the process is:
[tex]n_{tot} = \frac{1}{2}\cdot \ddot n_{1}\cdot t_{1}^{2} + \dot n_{1}\cdot t_{2}+\frac{1}{2}\cdot \ddot n_{2}\cdot t_{2}^{2}[/tex]
[tex]n_{tot} = \frac{1}{2}\cdot \left(\frac{2}{3}\,\frac{rev}{s^{2}} \right)\cdot (9\,s)^{2} + \left(6\,\frac{rev}{s}\right)\cdot (13\,s) + \frac{1}{2}\cdot \left(-\frac{6}{13}\,\frac{rev}{s^{2}} \right) \cdot (13\,s)^{2}[/tex]
[tex]n_{tot} = 66\,rev[/tex]
