Question

Where should I start?

Answer 1

centripetal force = tension

Answer 2

ie mv^2/r=39 N
m*14*14/2.8=39

Answer 3

m=39/70;

Answer 4

I tried this answer; incorrect.

Answer 5

actually that's not right for a vertical path...

Answer 6

decompose mg into radial and tangential components...

Answer 7

you'll see that one component of mg provides some centripetal force...
so T will be less than mV^2/r

Answer 8

Should I break the weight of the object into vertical and horizontal component? (x - direction is parallel to the direction of tension)

Answer 9

not horizontal and vertical... for rotational problems the convenient axes are radial and tangential:

Answer 10

Isn't it the same as horizontal and vertical component?

Answer 11

no.

Answer 12

mg is vertical

Answer 13

If you let x direction be parallel to tension, then it is kind of the same thing.

Answer 14

can u tel me is this correct @Algebraic! mg+mv^2/r=T

Answer 15

@ashishthomas7 no..

Answer 16

Like this:

Answer 17

@micahwood50 best to call it radial and tangential.... horizontal is something else entirely...

Answer 18

sure.

Answer 19

I can't see the difference.

Answer 20

or cut out the needless extra labels and just call it radial and tangential...

Answer 21

since those are useful concepts in rotational motion...

Answer 22

eg. V tangential = omega*r etc...

Answer 23

So \[\Large F_c = T + Wsin \theta = m a_c\]

Answer 24

a radial = (V tangential)^2 /r and so on...

Answer 25

Sorry, I don't know radial and tangential...

Answer 26

just a components.