a stone of mass 0.5kg tied with a string of length 1m is moving in a horizontal circle with speed 4m/s.the tension in the string is...

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a stone of mass 0.5kg tied with a string of length 1m is moving in a horizontal circle with speed 4m/s.the tension in the string is...

Physics
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|dw:1327670211788:dw| for vertical equilibrium
hey salini
heloo

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Other answers:

endha answer ariyo aravind?
maybe
tht answer to my qn was wnderful i was looking for tht
endha?
?
u gud in malayalam??
ene kurichu endha vijaricha ney?
enaal nallathu in this qn enikku thonunu tension is provided by the centripetal force
helo
annu pakshe e questionil angle kurichu onum thannela aa diagram yaane varachadhannu
y do u need angle for mv^2/r
|dw:1327671459182:dw|alla imagine chei if there was no angle then there wud be no force in the vertical direction to balance mg
http://www.twiddla.com/735087
The centripetal force, \( F_c \) required to keep the object in uniform circular motion is \[ F_c = \frac{mv^2}{r} \] |dw:1327674336876:dw| F_c is the force pointing towards the center of rotation. Now, where is that F_c provided from? It is provided but the tension, T, in the string. It is the component in F_c in that direction. The other piece of T is the vertical component and you know that must be equal to the weight of the mass, mg. If the string makes an angle theta to the vertical, we must have therefore that \[ T\cos\theta = mg \ \ \ \ -- (1) \] \[ T \sin\theta = F_c = \frac{mv^2}{r} \ \ \ \ -- (2) \] Now you know the length of the string, it is 1 m. And you know the radius of the rotation, it is \( r \). Hence another expression involving \( \theta \) is \[ \sin\theta = \frac{r}{1} = r \ \ \ \ -- (3) \] You now have enough equations to be able to solve for \( T \).
always ina horizontal circle there is some angle made with vertical or horizontal right?
By the string, yes.
:P
hey
u thr?

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