Respuesta :
Answer:
69 revolutions
Explanation:
given,
initial angular velocity = 0
final angular velocity = 6 rev/s
when safety switch turned off
tub smoothly slows to rest in (t)= 13 s
total revolution of tub = ?
angular acceleration = [tex]\dfrac{\omega_f-\omega_0}{t}[/tex]
angular acceleration = [tex]\dfrac{6-0}{10}[/tex]
= 0.6 rev/s²
now,
[tex]\theta = \omega_i t +\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta = 0+\dfrac{1}{2}\times 0.6 \times 10^2[/tex]
= 30 revolution
now,
when safety switch turn off
angular acceleration = [tex]\dfrac{\omega_f-\omega_0}{t}[/tex]
angular acceleration = [tex]\dfrac{0-6}{13}[/tex]
= -0.462 rev/s²
now,
[tex]\theta = \omega_i t +\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta = 6 \times 13 -\dfrac{1}{2}\times 0.462 \times 13^2[/tex]
= 39 revolution
total revolution = 30 + 39 = 69 revolutions.
tub will rotate 69 revolutions
