A constant frictional torque of 220 Nm is applied to a turbine initially rotatin

Question

A constant frictional torque of 220 Nm is applied to a turbine initially rotating at 20 rev/min, and it comes to a stop in 20 s. What is turbine's moment of inertia? What is the total work done by frictional forces? A particle executes linear harmonic motion about the point x = 0. At t = 0 it has displacement x = 0.6 cm and zero velocity. The frequency of the motion is 1.5 Hz. Determine (a) the period; (b) the angular frequency; (c) the amplitude; (d) the maximum speed; (e) the maximum acceleration; (f) the displacement at t = 4.5 s, and (g) the speed at t = 4.5 s. The propeller of a light plane has a length of 1.8192 m and rotates at 2160 rpm. The rotational kinetic energy of the propeller is 124.3 kJ. What is the mass of the propeller? You can treat the propeller as a thin rod rotating about its center.

Explanation / Answer

1) Torque = -220 N.m ,

wO = 20 rev/min = 20*2*3.14/60 = 2.093 rad/s

w =0, t = 20 s

alpha = (w-wo)/t

alpha = (0-2.093)/20 = -0.10465 rad/s^2

Torque = I*alpha

-220 = I*-0.10465

I 2102.24 kg.m^2

2) x = 0.6 cm , f = 1.5 Hz

(a) T = 1/f = 1/1.5 = 0.67 s

(b) w = 2pi*f = 2*3.14*1.5 = 9.42 rad/s

(c) A =0.6 cm

(d) vmax = Aw = 0.6*9.42

vmax = 5.65 cm/s

(e) amax = Aw^2 = 0.6*9.42^2

amax = 53.2 cm/s^2

(f) t = 4.5 s

x = Acos(wt)

x = 0.6cos(9.42*4.5)

x = -0.013 cm

(g) v = - Awsin(wt)

v = -(5.65)sin(9.42*4.5)

v = 5.6487 cm/s

3) moment of inertia of rod about its center I = mL^2/12

L = 1.8192 m

w = 2160 rpm = 2160*2*3.14/60 = 226.08 rad/s

W = 124.3 kJ

W = 0.5Iw^2

124.3*1000 = 0.5*m*1.8192^2*226.08^2/12

m = 17.64 kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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