Respuesta :
Answer:
The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions
Explanation:
Given data
[tex]\omega_1[/tex] = 0
[tex]\omega_2[/tex] = 5 [tex]\frac{rev}{sec}[/tex]
Time taken = 7 sec
(1). The angular acceleration is given by
[tex]\alpha = \frac{d \omega}{dt}[/tex]
[tex]\alpha = \frac{5}{7}[/tex]
[tex]\alpha = 0.714 \frac{rev}{s^{2} }[/tex]
We know that from the equation of motion
[tex]\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1[/tex]
[tex]5^{2} = 0 + 2 (0.714) \theta_1[/tex]
[tex]\theta_1 = 17.5 \ rev[/tex] -------- (1)
(2). The angular acceleration is given by
[tex]\alpha = \frac{d \omega}{dt}[/tex]
[tex]\alpha = - \frac{5}{14}[/tex]
[tex]\alpha =[/tex] - 0.357 [tex]\frac{rev}{s^{2} }[/tex]
We know that from the equation of motion
[tex]\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2[/tex]
[tex]0^{2} = 5^{2} + 2 (-0.357) \theta_2[/tex]
[tex]\theta_2[/tex] = 35.01 rev ------- (2)
Total no of revolution made by the machine is
[tex]\theta = \theta_1 + \theta_2[/tex]
[tex]\theta =[/tex] 17.5 + 35.01
[tex]\theta =[/tex] 52.51 rev
Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev
