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Finding Velocity

  • Thread starter joemama69
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  • #1
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Homework Statement



The soap-box car has a weight of 110 lb, including the passenger but excluding its four wheels. Each wheel has a weight of 5 lb, radius of .5 ft, and a radius of gyration k = .3 ft, computed about an axis passing through the wheels axle. Determine the cars speed after it has traveled 100 ft starting from rest. the wheels roll without slipping. Neglect Air.


Homework Equations



T1 = [tex]\sum[/tex]U1-2 = T2


The Attempt at a Solution



T1 = 0 because it starts from rest
for T2 the mass would be 110lb car pluss the 4 5lb wheels, converted to slugs of course, and the v is what i am looking for

im not sure about the middle term

I = mk3 = .1554*.32 = .013986

What else do i need
 

Answers and Replies

  • #2
LowlyPion
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What force is acting on the soapbox?

Incline? Angle?
 
  • #3
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Fy = 130 lb
 
  • #4
LowlyPion
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  • #5
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T2 = .5mv2 + .5Iw2

=.5(130/32.2)v2 + .5(130/32.2).32w2

T2 = 325 / 161 v2 + 117 / 644 w2

v = .5w2

T2 = 221/322 w2

[tex]\sum[/tex]U1-2 = V1 - V2, V = Wy

V1 = Wy = W(100sin30)

V2= 0

Therefore

100Wcos30 = 221 / 322 w2

Something aint right
 
  • #6
LowlyPion
Homework Helper
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Something aint right
First they have given you the weights. You need the mass.

Second what is sin30? Is that the angle I previously inquired about being given?
 
  • #7
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oops sorry, i didnt understand when u asked the incline angle

it is 30 degrees

and they hypotenus is 100ft, so 100sin30 = y = 50

&

I did convert to mass when i divided the weight 130 by 32.2

Also...

When do use the info on the wheels, r = .5, k = .3, w = 5 lb
 
  • #8
LowlyPion
Homework Helper
3,090
4
oops sorry, i didnt understand when u asked the incline angle

it is 30 degrees

and they hypotenus is 100ft, so 100sin30 = y = 50

&

I did convert to mass when i divided the weight 130 by 32.2

Also...

When do use the info on the wheels, r = .5, k = .3, w = 5 lb
Oh, I see. You got it. I see the 32.2 now.

OK the radius of gyration basically gives you the I = m*Rg2

But for your linear translational kinetic energy you have a r = .5 ft
 
  • #9
399
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I = mk2 = (5/32.2).32 = 9/644 <-- Should i multiply this my 4 since there are 4 wheels

50W = 1/2(130/32.2)v2 + 1/2 (9/644)w2

w = v/.5

50W = (325/161)v2 + (9/1288)(v/.5)2

50W = 325/161 v2 + 9/322 v2

50w = 659 / 322 v2

v = (16100W) / 659 or v = 24.43095599W

Is this right so far, How do i find W
 
  • #10
399
0
I = mk2 = (5/32.2).32 = 9/644 <-- Should i multiply this my 4 since there are 4 wheels

50W = 1/2(130/32.2)v2 + 1/2 (9/644)w2

w = v/.5

50W = (325/161)v2 + (9/1288)(v/.5)2

50W = 325/161 v2 + 9/322 v2

50w = 659 / 322 v2

v = (16100W) / 659 or v = 24.43095599W

Is this right so far, How do i find W
 
  • #11
399
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brain fart

W = 130 right

so v = 56.35622 ft/s
 
  • #12
399
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anyone agree, disagree
 
  • #13
LowlyPion
Homework Helper
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anyone agree, disagree
m*g*h = 50*130 = 6500 = 1/2*m*V2 + 1/2*I*ω2

I = 4*5/32.2*(.3)2 = .056

ω = V/.5

6500 = 1/2*(4.037 + .056)*V2

V = 56.36

Looks OK.
 

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