Calculating Angular Speed of Amusement Park Ride

[theNeonGod]

angular speed please help


i just don't see how i can get this w/out knowing anything containing time:

in an amusement park rocket ride, cars are suspeded from 4.25 m cables attached to rotating arms at a distan o 6 m from the axis of rotation. The ables swing out at an angle of 45 degrees when the ride is operating. What is the angular speed of rotation?

any help would be more than appreciated. I'm sort of desperate at this point.

thank you.

--
eric

 

[physics noob]

i would use the equation Ac = V^2 / r multiply that by distance to get work done,,,, then you would have to incorporate gravity , oh and there would be a tension toward the middle of the circle... I am not really sure the specifics of your problem as it is hard to understand, but the steps above is what you would do to solve it, you would neeed to use them.

 

[theNeonGod]

got it. awesome. thanx.

 

[siddharth]

First of all, draw the Free Body diagram for one car in the ride.
From the frame of the car, there is a centrifugal force acting towards the center.

Also, gravity and Tension are acting.
Now, let the y-axis be in the vertical direction and x-axis in the horizontal direction.
What is the component of the Tension in the y-direction? This component must be equal to the force due to gravity as there is no acceleration. From this you can find the Tension in the rope.
Also, the component of Tension in the x-direction must be equal to the centrifugal force, which is (mv^2)/(r+R) {Where r + R is the horizontal distance from the axis of rotation(Use some trignometry to find it). Here, the car is not tied directly to the axis of rotation}. And you know that v=(R+r)w. From that you can find angular speed.