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anonymous
 4 years ago
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...
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327670211788:dw for vertical equilibrium

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0endha answer ariyo aravind?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0tht answer to my qn was wnderful i was looking for tht

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ene kurichu endha vijaricha ney?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0enaal nallathu in this qn enikku thonunu tension is provided by the centripetal force

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0annu pakshe e questionil angle kurichu onum thannela aa diagram yaane varachadhannu

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0y do u need angle for mv^2/r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327671459182:dwalla imagine chei if there was no angle then there wud be no force in the vertical direction to balance mg

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2The 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 \( T \).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0always ina horizontal circle there is some angle made with vertical or horizontal right?
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