Attached to each end of a thin steel rod of length 1.90 m and mass 6.80 kg is a

Question

Attached to each end of a thin steel rod of length 1.90 m and mass 6.80 kg is a small ball of mass 1.10 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 40.0 rev/s. Because of friction, it slows to a stop in 36.0 s. Assuming a constant retarding torque due to friction, compute (a)the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the 36.0 s

Explanation / Answer

Here,

l = 1.90m
r = 1.90/2 = 0.95 m
m1 = 6.80 kg
m2 = 1.10 kg
w = 40 rev/s = 251.327 rad/s

Moment of inertia = 0.5mr^2 + 0.5mr^2 + 1/12 * Mr*l^2
I = 0.5*1.10*0.95^ + 0.5*1.10*0.95^ + 0.0833*6.80*1.90^2
I = 14.414 kg.m^2

wf = wi + at
a = wf/t
a = 251.327/36
a = 6.981 rad/s^2

Torque = I*a
WHere,
a is angular acceleration

T = 14.414 * 6.981
T = 100.62

Total Energy = 0.5 * I * w^2
E = 0.5 * 14.414 * 251.327^2
E = 455232.035 J
E = 455.232 KJ

Angular distance = 0.5 * a *t^2
AD = 0.5 * 6.981 * 36^2
AD = 4523.688 rad
AD = 4523.688 * (1/2pi)
AD = 719.967 rev/s

No of revolution = 719.976/(2*3.14)
n = 114.64

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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