Kinetic energy and Angular momentum

[Stryder_SW]

Homework Statement

Five identical particles of mass m = 0.30 kg are mounted at equal intervals on a thin rod of L = 0.87 m and mass M = 2.0 kg, with one mass at each end of the rod. If the system is rotated with angular velocity ω = 50 rev/min about an axis perpendicular to the rod through one of the end masses, determine the following.

(a) the kinetic energy of the system

(b) the angular momentum of the system


2. The attempt at a solution
SOLVED
(a) K=.5(I_total)w^2
K=.5(1/3L²*M + 30/16L²*m)w²
K=12.75312242kg·(m/s)²
(b)Angular Momentum=I_total*w
Angular Momentum=4.87133934kg·m²/s

 

[LowlyPion]

They want you to develop the I for this configuration.

If there are 5 evenly spaced, then you have to determine the distances (L/4 is the separation) and then develop the sum of the moments of inertia of the parts. The rod and the weights. However this is a little trickier. The weight in the middle needs to be properly accounted for.

So you can use the I of a rod rotated about it's end and then the 4 weights evenly spaced at L/4 along its length.

 

[Stryder_SW]

Well what you're describing pretty much sounds like what I did, Find the I of the rod which is just (1/3)Md^2 where d = .87m. Then find the Inertia of each particle with a varying L which would come out as L1=0, L2=.2175, L3=.435,L4=.6525,L5=.87. so the general form for the Inertia of the particles is .5m*(Ln)^2 (n being 1-5). then just add all Irod and Iparts. multiply by (w^2)/2.

 

[LowlyPion]

Well what you're describing pretty much sounds like what I did, Find the I of the rod which is just (1/3)Md^2 where d = .87m. Then find the Inertia of each particle with a varying L which would come out as L1=0, L2=.2175, L3=.435,L4=.6525,L5=.87. so the general form for the Inertia of the particles is .5m*(Ln)^2 (n being 1-5). then just add all Irod and Iparts. multiply by w^2.

From the end I get (1/3L²*M + 30/16L²*m) for I

 

[Stryder_SW]

what do you mean by "from the end"

 

[LowlyPion]

what do you mean by "from the end"

about an axis perpendicular to the rod through one of the end masses...

From the problem right? Don't you need I relative to one end?

 

[Stryder_SW]

yeah, relative to the end its rotating about. but is that I equation you gave supposed to be the total I? more specifically is the (30/16L2*m) part the I for all the particles summed together? and if it was all the particles added together what would L be?

 

[LowlyPion]

yeah, relative to the end its rotating about. but is that I equation you gave supposed to be the total I? more specifically is the (30/16L2*m) part the I for all the particles summed together? and if it was all the particles added together what would L be?

m is the mass of the little weights.

(L/4)² + (L/2)² + (3L/4)² + L² = 30/16 L²

1/16 + 4/16 + 9/16 + 16/16 = 30/16 right?

 

[Stryder_SW]

ah, I see. That makes so much more sense now XD. so I_total=(1/3L²*M + 30/16L²*m), so to get K it would be (1/2)(I_total)w² and in this case w=50rev/min

so I_total=(1/3(.87m)²*2kg + 30/16(.87m)²*.3kg)=.93035625kg*m²
K=(1/2)(.93035625kg*m²)(5.235987756/s))²=12.75312242 kg*(m²/s²)
OK got it, thank you SO MUCH Pion.

 

[LowlyPion]

ah, I see. That makes so much more sense now XD. so I_total=(1/3L²*M + 30/16L²*m), so to get K it would be (1/2)(I_total)w² and in this case w=50rev/min

so I_total=(1/3(.87m)²*2kg + 30/16(.87m)²*.3kg)=.93035625kg*m²
K=(1/2)(.93035625kg*m²)(50/60(rev/s))²=.323043646 kg*(m²/s²)
Obviously I'm still doing something wrong, cause that's not a right answer.

There you go. You want radians/sec.

 

[LowlyPion]

ah, I see. That makes so much more sense now XD. so I_total=(1/3L²*M + 30/16L²*m), so to get K it would be (1/2)(I_total)w² and in this case w=50rev/min

so I_total=(1/3(.87m)²*2kg + 30/16(.87m)²*.3kg)=.93035625kg*m²
K=(1/2)(.93035625kg*m²)(5.235987756/s))²=.323043646 kg*(m²/s²)
OK got it, thank you SO MUCH Pion.

Much better.

Good luck.