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# M3 Circular Motion watch

1. A light elastic string AB has nat. length l and modulus (20mg)/3 the end A is fixed and a particle of mass m is attached to end B

the particle is made to move in a horizontal circle at constant angular speed with the string inclined at an angle arcsin(4/5) to eh vertical. Find in terms of l and g the angular speed of the particle.

the answer is 2(g /3l )^1/2
(i.e 2 times the sq rt of g over 3l)
2. Resolve upwards: TcosQ = mg (Q=theta)

sinQ = 4/5, so cosQ = 3/5
so, 3T/5 = mg

T=(lambda)(e)/L = (20mg/3)(e) / L

(3/5)(20mg/3)(e) / L = mg

e=L/4
so length of string = 5L/4

F=ma
TsinQ = mr(w^2)
eventually: w^2 = 4(g/3L)

so w = 2(g/3L)^1/2
3. wat does r equal though? cuz i got uptil that point but then i cudnt find the radius r
4. (Original post by chetan)
wat does r equal though? cuz i got uptil that point but then i cudnt find the radius r
sinQ = r/(5L/4) = 4r/5L

so you get T(4r/5L) = mr(w^2)
now the r's can cancel, and so on.....
hope it makes sense now.
5. ye thats wat i 4got. thanx a lot

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