SOLUTION: I know this is not algebra but if you can answer that would be great. A uniform rod with a 3.0 m length and a mass of 14.0 kg is supported by a hinge at its left end and held in

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Question 1831: I know this is not algebra but if you can answer that would be great.
A uniform rod with a 3.0 m length and a mass of 14.0 kg is supported by a hinge at its left end and held in a horizontal position. A mass of 6.0 kg is attached to the rod 0.36 m from the hinge. A second mass of 7.5 kg is attached to the rod 2.80 m from the hinge. (A) Find the moments of inertia for each of the three masses about an axis at the hinge. (B) At the instant the right end of the rod is released, what is the net torque acting on this system? (C) What is the angular acceleration of this system the instant it is released?
Found 2 solutions by khwang, Ne0:
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
A uniform rod with a 3.0 m length and a mass
of 14.0 kg is supported by a hinge at its left end and held in a
horizontal position. A mass of 6.0 kg is attached to the rod 0.36 m
from the hinge. A second mass of 7.5 kg is attached to the rod 2.80 m
from the hinge. (A) Find the moments of inertia for each of the three
masses about an axis at the hinge. (B) At the instant the right end of
the rod is released, what is the net torque acting on this system?
(C) What is the angular acceleration of this system the instant it is released?
Sol: (a) Center of mass of the rod is 3/2 = 1.5 m from the hinge,
The moments of inertia of the rod about the hinge = 14* (1.5)^2 = 31.5 kg*m^2
The moments of inertia of the mass- 6 kg = 6* (0.36)^2 = 0.7776 kg*m^2
The moments of inertia of the mass- 7.5 kg = 7.5* (0.28)^2 = 0.588 kg*m^2
(Total of 3 moments of inertia = 32.8656 kg*m^2)
(b) Net Torque = r x F = 1.5 * 14 g + 0.36* 6 g + 0.28* 7.5g
= (1.5 * 14 + 0.36* 6 + 0.28* 7.5)*9.8 = 247.548 meter X kg * meter/sec^2
(c) The angular acceleration = Net Torque/ (Sum of The moments of inertia in (a))
= 247.548/32.8656 = 7.53 (radians/ sec)

This is a physics problem not in the category of algebra.
Kenny

Answer by Ne0(6) About Me  (Show Source):
You can put this solution on YOUR website!
(A) To find the inertia for any body is I=Icm+Mh^2. The center of mass in this case for the whole system is Xcm=%28m1X1%2Bm2X2%29%2F%28m1%2Bm2%29. Xcm=1.72 m from the hinge. The inertia of the center of mass is the inertia of the rod at the center of mass which is %28mr%5E2%29%2F%2812%29. The r is the distance we just got and m is the mass of the rod itself. Icm=3.43 kgm^2. If M is the total mass of the system while h is the distance at the center of mass the total inertia you get is 3.43+(6+7.5+14)(1.72)^2=84.37 kgm^2.

(B) The net torque is T=rF. Since the force is applied tangentially you don't have to worry about the sin of the angle between the radius and the force. The Force is simply the total mass of the system multiplied by 9.8 with which we get (6+7.5+14)(9.8)=269.5 N. The radius is once again the distance from the center of mass which is 1.72m so the net torque is T=(1.72)(269.5)=463.54 N*m.

(C) Remember the other definition of torque is that it is equall to the inertia mutiplied by the angular acceleration. This means that the angular acceleration is equivalent to the net torque divided by the inertia of the system or %28T%29%2F%28I%29. Angular acceleration=%28463.54%29%2F%2884.37%29=5.49 rad/sec^2.