Item 4 Part A A high-speed flywheel in a motor is spinning at 450 rpm whena powe

Question

Item 4 Part A A high-speed flywheel in a motor is spinning at 450 rpm whena power failure suddenly occurs. The flywheel has mass 36.0 kg and diameter 72.0 cm. The power is off for 33.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the fiywheel makes 160 complete revolutions. At what rate is the flywheel spinning when the power comes back on? rad/s Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part B How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? Submit My Answers Give Up Part C How many revolutions would the wheel have made during this time? rev Submit My Answers Give Up Provide Feec

Explanation / Answer

Moment of inertia of flywheel is I = 0.5*m*r^2 = 0.5*36*(0.72/2)^2 = 2.33 Kg-m^2

initial angular velocity is wi = 450 rpm = 450*(2*3.142/60)= 47.13 rad/s

power cut ocuurs for t = 33 sec

angular displacement is theta = 160*2*3.142 = 1005.44 rad

angular accelaration is alpha

using kinematic equations

theta = (wi*t)+(0.5*alpha*t^2)

1005.44 = (47.13*33)+(0.5*alpha*33^2)

alpha = -1 rad/s^2

then also using

W = (wo)+(alpha*t)

w = 47.13-(1*33) = 14.13 rad/sec

-------------------------------------------------

B) finally to stop the flywheel time taken is t

angular velocity is w = 0 rad/s

wi = 47.13 rad/s

alpha = -1 rad/s^2

t = ?

using

W = wi+(alpha*t)

0 = (47.13)-(1*t)

t = 47.13 sec

-------------------------------------------------------------

using theta = (wi*t)+(0.5*alpha*t^2)

theta = (47.13*47.13)-(0.5*1*47.13^2)

theta = 1110.6 rad

then required no.of revolutions are N = 1110.6/(2*3.142) = 176 rev

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.