# Incline and Rotational Energy/Work

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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∑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

tiny-tim
Homework Helper
Hi Torater! sin30 (5a) + 8a = 39.24

a= 6.04m/s²
uhh? 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:
tiny tim i did my algebra wrong!! 3.74??

Last edited:
tiny-tim
Homework Helper
M=5+cos30(5)
mt=7.5 + 8
I'm sorry, but you've gone totally beserk Try again! 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

tiny-tim
Homework Helper
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? (and your arithmetic in your "long approach" was wrong, so it's difficult to check the rest)

(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?

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

tiny-tim
Homework Helper
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! (you really do need as many checks as possible, don't you? )

so then my tension should work out to 30.2..... haha lets hope I did this right

tiny-tim
Homework Helper
so then my tension should work out to 30.2
No … how did you get that? T= 8(9.81sin30 - 3.02)
T= 15.1N?

Before I did this:
T=8(9.81sin30 - 3.02)/sin30

tiny-tim
Yup! (using the short-cut, you could also have said it was 5 times 3.02 = 15.1 )