Tension of an Object Swinging in a Circle

[wallace13]

A "swing" ride at a carnival consists of chairs that are swung in a circle with a 10 m radius. It is attached to a vertical rotating pole with a 56.5 degree angle between the pole and chain of a swing. Suppose the total mass of a chair and its occupant is 210 kg.

(a) Determine the speed of the chair.
I found the speed of the chair by using the equation tan 0 = m x v squared/gr
I got 12.175m/s and it is correct.

(b) Determine the tension in the cable attached to the chair.
I plugged the velocity that I found into the equation Fc= m x v squared/r
I got 3112 N, which is incorrect



tan 0= m x v squared/gr
Ac= v squared/r
Tc= mxac

 

[Doc Al]

(b) Determine the tension in the cable attached to the chair.
I plugged the velocity that I found into the equation Fc= m x v squared/r

That will give you the centripetal force, not the tension in the cable. Hint: Consider the components of the cable tension. (You would be wise to draw a free body diagram illustrating the forces acting on the chair.)

 

[thomate1]

= mv squared/gr

In order to determine the tension in the cable, we need to calculate the centripetal acceleration (Ac) of the chair. This can be done by using the equation Ac = v^2/r, where v is the velocity and r is the radius of the circle.

Plugging in the values, we get Ac = (12.175 m/s)^2 / 10 m = 14.84 m/s^2.

Now, we can use this value of acceleration to calculate the tension in the cable using the equation Tc = m x Ac, where m is the mass of the chair and occupant.

Tc = (210 kg) x (14.84 m/s^2) = 3116.4 N.

Therefore, the tension in the cable attached to the chair is approximately 3116.4 N. It is important to note that this is the minimum tension required to keep the chair and occupant moving in a circular motion. Any lower tension would result in the chair and occupant flying off the ride.