More circular motion questions:

[jiboom]

these are only short queries so i put them in the same thread.

1) one end of a light inelastic string of length a is attached to a fixed point A and a particle of mass m is attached to the other end,B. The particle is held at the same level as A,at a distance of asin x away from A,and released. find
a) the impulse in the string when it becomes tau
b) the speed of the particle immediately after the string becomes taut
c) the cosine of the angle between the string and the vertical when the particle first comes to instantaneous rest.

for the last part i can get the answer only if i know the particle goes above my PE=0 line which i chose to be at the point of the jerk. is this obviously the case?

3)
a particle P of mass 2m is attached by a light inextensible string of length a to a fixed point O and is also attached by another light inextensible string of length a to a small ring Q of mass 3m which can slide on a fixed smooth vertical wire passing through O. the particle P describes a horizontal circle with OP inclined at an angle 60 with the downward vertical.

a) find the tensions
b) show speed of P is (6ga)^1/2
c) find the period of revolution of the system

how do i find part c)? i have used v to find w (angular velocity) but this does not give me the book answer.the radius of the circle being traveled is art(3)/2 if it helps.

 

[jiboom]

anyone able to tell me how to find period?

 

[jiboom]

a sly bump as the thread asking how to find a in f=ma has had more replies :0

surely someone can tell me how to find the period of the system given the speed of
v=(6ga)^1/2

the answer,if it helps, is 2pi(a/6g)^1/2

 

[jiboom]

these are only short queries so i put them in the same thread.

1) one end of a light inelastic string of length a is attached to a fixed point A and a particle of mass m is attached to the other end,B. The particle is held at the same level as A,at a distance of asin x away from A,and released. find
a) the impulse in the string when it becomes tau
b) the speed of the particle immediately after the string becomes taut
c) the cosine of the angle between the string and the vertical when the particle first comes to instantaneous rest.

for the last part i can get the answer only if i know the particle goes above my PE=0 line which i chose to be at the point of the jerk. is this obviously the case? .

have been assured my answer is correct for question 2,book is wrong. i have only the last part of question 1. is there an obvious reason the particle goes aboive my pe=0 line?

 

[asteriatic]

1) a) The impulse in the string can be found using the formula J = m(v-u), where m is the mass of the particle, v is the final speed and u is the initial speed. In this case, u = 0 as the particle is initially at rest. The final speed can be found using the formula v^2 = u^2 + 2aΔx, where a is the acceleration and Δx is the distance the particle travels. In this case, a = g (acceleration due to gravity) and Δx = a sin x. Plugging in the values, we get v = √(2ga sin x). Therefore, the impulse in the string is J = m(√(2ga sin x) - 0) = m√(2ga sin x).

b) The speed of the particle immediately after the string becomes taut is also given by v = √(2ga sin x).

c) To find the cosine of the angle between the string and the vertical, we can use the formula cosθ = (v^2 - u^2)/(2ga), where θ is the angle between the string and the vertical. Plugging in the values, we get cosθ = (√(2ga sin x)^2 - 0^2)/(2ga) = sin x. Therefore, the cosine of the angle is equal to sin x.

2) It is not necessary that the particle goes above the PE=0 line. The angle x can be any value and the calculations will still hold true.

3) a) The tension in the strings can be found by considering the forces acting on the particle P. There is the weight of the particle acting downwards, the tension in the string attached to Q acting upwards and the tension in the string attached to O acting at an angle of 60 degrees to the horizontal. Using trigonometry, we can find the tension in both strings to be T = mg/sin 30 = 2mg.

b) To show that the speed of P is (6ga)^1/2, we can use the formula v^2 = u^2 + 2aΔx, where u is the initial speed (0 in this case), a is the acceleration (in this case, the centripetal acceleration) and Δx is the distance traveled. The distance traveled can be