Physics problem! Please help!

Question

anonymous · Answer

attachment

ujjwal · Answer

Actually$$T=\frac{mv^2}{r}+mg\cos\theta$$

anonymous · Answer

So far I have Fc=9m/2, this accounts for the tension due to the rotation.  I don't know how to incorporate gravity into it without the angle.

ujjwal · Answer

and$$mg\sin\theta=ma$$
Find $\theta$ from this relation and replace it in my above relation!
you will get m

anonymous · Answer

I just don't know how to find θ, I think that's the only problem...

anonymous · Answer

@derrick902  wat should be the answer...)

anonymous · Answer

The answer is 2kg

ujjwal · Answer

If you use the process suggested by me, you will get 2.2 kg

anonymous · Answer

I agree, T=mv2r+mgcosθ is the right formula, but how to find theta!

anonymous · Answer

@ujjwal  why  adding mgcosx

ujjwal · Answer

I used $mg\sin\theta=ma$

anonymous · Answer

mgcosx would be the additional tension due to gravity on the mass

ujjwal · Answer

@Yahoo! , because $mg\cos\theta$ is balanced by tension in string!

shubhamsrg · Answer

|dw:1350487807613:dw|
so 
mgcos@ =T- mv^2 /r
and
mgsin@ = ma_r

square and add both..

@ dissapears..

you also now  a_net = sqrt( a_c^2 + a_r^2 )


that should do it maybe..

shubhamsrg · Answer

a_c = centripetal accn
a_r = radial accn

shubhamsrg · Answer

am sorry its not radial but it should be tangential accn a_t in that..everything is same,,just replace a_r by a_t ..

anonymous · Answer

I think I understand most of it what you said, but how did you get a_t without theta?

anonymous · Answer

you don't,  you use that equation to find theta..

anonymous · Answer

$$7 = \sqrt{a _{t}^2+a _{c}^{2}}$$