A uniform rod of mass 1.70 kg and length 2.00 m is capable of rotating about an

Question

A uniform rod of mass 1.70 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 4.90 kg is attached to one end and a second mass m2 = 2.00 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?kg · m2
(b) If the rod rotates with an angular speed of 2.60 rad/s, how much kinetic energy does the system have? J
(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined? kg · m2
(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.60 rad/s? J

Explanation / Answer

Given

uniform rod of mass M = 1.70 kg, length is 2 m

and m1= 4.90 kg, m2 = 2.00 kg are at ends of the rod

we know that the moment of inertia of the rod rotating about the axis passing through the centre and perpendicular to the length is I = 1/12 Ml^2

a)
   so moemtnt of inertia of Rod is I = (1/12)(1.782^2) kg m2 = 0.264627 kg m2

now the moment of inertia of the point masses at the ends is

   I1 = m1(l/2)^2 = 4.9(2/2)^2 = 4.9 kg m2
and

   I2 = m2(l/2)^2 = 2(2/2)^2 = 2 kg m2

total moment of inertial of the system is = I+I1+I2 = 0.264627+4.9+2 = 7.164627 kg m2

b) rotational kinetic energy is k.e = 0.5*I*W^2

   W = 2.60 rad/s

               k.e = 0.5*7.164627*2.6^2 J

               k.e = 24.21643926 J

c) considering the mass of the rod negligible then

   the moment of inertia of the two mass system is located at x from the m1

   m1--------x------------m2

   <---x-----><-----2-x---->

equating the momets about the centre of mass then

   4.9*x*g = 2(2-x)g

   4.9x = 4-2x
   x = 0.57971 m
and 2-x = 2-0.57971 m = 1.42029 m

now the moment of inertial is = m1x^2+m2(2-x)^2

           = 4.9*0.57971^2+2*1.42029^2 kg m2 = 5.6812 kg m2

d) w = 2.60 rad/s then rotational kientic energy k.e = 0.5*I*W^2

           k.e = 0.5*5.6812*2.6^2 J = 19.202456 J

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