## 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...

1. anonymous

|dw:1327670211788:dw| for vertical equilibrium

2. AravindG

hey salini

3. AravindG

heloo

4. anonymous

endha answer ariyo aravind?

5. AravindG

maybe

6. AravindG

tht answer to my qn was wnderful i was looking for tht

7. anonymous

endha?

8. AravindG

?

9. AravindG

u gud in malayalam??

10. anonymous

ene kurichu endha vijaricha ney?

11. AravindG

enaal nallathu in this qn enikku thonunu tension is provided by the centripetal force

12. AravindG

helo

13. anonymous

annu pakshe e questionil angle kurichu onum thannela aa diagram yaane varachadhannu

14. AravindG

y do u need angle for mv^2/r

15. anonymous

|dw:1327671459182:dw|alla imagine chei if there was no angle then there wud be no force in the vertical direction to balance mg

16. AravindG
17. JamesJ

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 $$T$$.

18. anonymous

always ina horizontal circle there is some angle made with vertical or horizontal right?

19. JamesJ

By the string, yes.

20. AravindG

:P

21. AravindG

hey

22. AravindG

u thr?