For the moment of inertia I got 276.28 and 21.25 which were both wrong and I can
ID: 2022090 • Letter: F
Question
For the moment of inertia I got 276.28 and 21.25 which were both wrong and I can't move on til I get them right. Any feedback would be greatly appreciated!!
Four particles with masses 2 kg, 4 kg, 2 kg,
and 5 kg are connected by rigid rods of neg-
ligible mass as shown. Assume the system
rotates in the yz plane about the x axis with
an angular speed of 2 rad/s.
(a) Find the moment of inertia of the system about the x axis.
Answer in units of kg · m2.
(b)Now assume the system rotates in the xy
plane about the z axis (origin, O) with an
angular speed of 2 rad/s. Find the moment of inertia of the system
about the z axis.
Answer in units of kg · m2.
(c) Find the rotational energy of the system
about the z axis.
Answer in units of J.
Explanation / Answer
For Part B you have to find the distance from each of the masses to the z axis using the pythagorean theorem. Because you know your side lengths, you can figure out that the diagonals (hypotnuses) are the lengths to the origin.
So your sides are 3.5 m (7/2), and 3m (6/2)
(3.5)2 + (3)2= 21.25
21.25= 4.6
Your distance to the origin from each mass (radius) is 4.6 m
Now you can use the equation for Inertia
I=mir2
and plug in the values for the masses so
(2kg)(4.6m)2 + (2kg)(4.6m)2 + (5kg)(4.6m)2 + (4kg)(4.6m)2
I=275.08 kgm2
For Part C:
The rotational kinetic energy is eaual to
Kr=1/2 I2
they already give you the value for omega as 2 rad/s, the angular velocity and you just found inertia in the last part
so plugging in the values
1/2(275.08 kgm2)(2 rad/s)2
Kr=550.16 J