Incline and Rotational Energy/Work

  • Thread starter Torater
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  • #1
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An 8kg mass is initally at rest on a 30 degree frictionless incline, it is attached to a hoop shaped 5kg pulley of radius 25.0cm by a massless cord. Find the tension in the cord and the acceleration of the mass using Newton's Laws.
 

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  • #2
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∑fy=m2a = N-m2gcos30
∑fx=-m2a = Tsin30-m2gsin30
∑τ= Iα = Tr

T= m2(gsin30-a)/sin30
I=mr²

m1a=m2(gsin30-a)/sin30

sin30 (5 a)= (8 (9.81sin30)-a)

sin30 (5a) + 8a = 39.24


a= 6.04m/s²

T= (8)(9.81sin30-6.04)/sin30
T= -18.2N
 
  • #3
tiny-tim
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Hi Torater! :smile:
sin30 (5a) + 8a = 39.24


a= 6.04m/s²
uhh? :confused:
 
  • #4
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r= .25m
m1=5kg
I=mr2
M=5+cos30(5)
mt=7.5 + 8
a= (8 (9.81)sin30/15.5

a= 2.53
 
Last edited:
  • #5
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tiny tim i did my algebra wrong!! 3.74??
 
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  • #6
tiny-tim
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M=5+cos30(5)
mt=7.5 + 8
I'm sorry, but you've gone totally beserk :redface:

Try again! :smile:
 
  • #7
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oh well thanks! haha the last 48hr's of physics frustration can do that to a person



ok attempt 2 at shortcut:

pseudo mass: mr^2 / r^2 = 5

M = 5+5 = 10

a= (8(9.81) sin30/10
a= 3.92


Ok lets focus on the long way as its what the question asks I don't think I did that right again.... maybe...

Or is my long approach still not right, and if not where? lol I got to start studying other chapters
 
  • #8
tiny-tim
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M = 5+5 = 10
No, the effective mass is the whole of the 8 plus the whole of the 5, = 13 …

whyever did you add the 5 to itself? :confused:

(and your arithmetic in your "long approach" was wrong, so it's difficult to check the rest)
 
  • #9
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(and your arithmetic in your "long approach" was wrong, so it's difficult to check the rest)

Where did I go wrong so I can correct this.... Like are my sum of force equations right?

Did I substitute them into other equations wrong? Or is the fact that I didn't use the Y forces at all wrong?


please direct me....

Whoops i didn't catch the whole effective mass part of that shortcut,
so it should come out to M=13 therefore a--> 8(9.81sin30)/13 = 3.02
 
  • #10
tiny-tim
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Where did I go wrong so I can correct this....
I showed you where, in post #3
Whoops i didn't catch the whole effective mass part of that shortcut,
so it should come out to M=13 therefore a--> 8(9.81sin30)/13 = 3.02
that's right! :biggrin:

(you really do need as many checks as possible, don't you? :wink:)
 
  • #11
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so then my tension should work out to 30.2..... haha lets hope I did this right
 
  • #12
tiny-tim
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so then my tension should work out to 30.2
No … how did you get that? :confused:
 
  • #13
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T= 8(9.81sin30 - 3.02)
T= 15.1N?



Before I did this:
T=8(9.81sin30 - 3.02)/sin30
 
  • #14
tiny-tim
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T= 8(9.81sin30 - 3.02)
T= 15.1N?
Yup! :biggrin:

(using the short-cut, you could also have said it was 5 times 3.02 = 15.1 :wink:)
 
  • #15
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Thank you so much for all your help, and putting up with my dumb errors throughout hahah
 

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