Incline and Rotational Energy/Work

[Torater]

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.

 

[Torater]

∑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]

Hi Torater!

sin30 (5a) + 8a = 39.24


a= 6.04m/s²

uhh?

 

[Torater]

r= .25m
m1=5kg
I=mr2
M=5+cos30(5)
mt=7.5 + 8
a= (8 (9.81)sin30/15.5

a= 2.53

 

[Torater]

tiny tim i did my algebra wrong! 3.74??

 

[tiny-tim]

M=5+cos30(5)
mt=7.5 + 8

I'm sorry, but you've gone totally beserk

Try again! ​

 

[Torater]

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 let's 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]

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)

 

[Torater]

(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

 

[tiny-tim]

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? )

 

[Torater]

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

 

[tiny-tim]

so then my tension should work out to 30.2

No … how did you get that?

 

[Torater]

T= 8(9.81sin30 - 3.02)
T= 15.1N?
Before I did this:
T=8(9.81sin30 - 3.02)/sin30

 

[tiny-tim]

T= 8(9.81sin30 - 3.02)
T= 15.1N?

Yup!

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

 

[Torater]

Thank you so much for all your help, and putting up with my dumb errors throughout hahah