Tension of an Object Swinging in a Circle

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The discussion centers on calculating the speed and tension of a chair on a carnival swing ride with a 10 m radius and a 56.5-degree angle. The speed of the chair was correctly calculated at 12.175 m/s using the tangent function. However, the initial calculation for the tension in the cable, which was found to be 3112 N, was incorrect. It was pointed out that this value represents centripetal force, not cable tension, and that a free body diagram should be used to analyze the forces acting on the chair. Understanding the components of cable tension is crucial for accurate calculations.
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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
 
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wallace13 said:
(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.)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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