In an amusement park ride, cars are suspended from L = 4.30 m cables attached to

Question

In an amusement park ride, cars are suspended from L = 4.30 m cables attached to rotating arms at a distance of D = 6.23 m from the axis of rotation. The cables swing out at an angle of ? = 53? when the ride is operating. What is the angular speed of rotation?

Explanation / Answer

Write equations from Newton’s 2nd law for the net forces on a rocket car in the horizontal and vertical directions. Horizontally, the force is centripetal, so: SF(x) = m?²r = Tsin?--------------------->(1) Vertically: SF(y) = 0 = Tcos?- mg mg = Tcos?-------------------------->(2) We don’t know T (tension in the cable) but we can eliminate it by dividing (1) by (2): m?²r/mg = Tsin?/Tcos? ?²r/g = tan? ? = v[gtan?/r] Now, r is going to be the length of the rotating arm plus the horizontal distance the car swings out, which is given by the product of the length of the cable and the sine of the angle made by the cable with the vertical: r = d + Lsin? So the speed formula above now looks like this: ? = v[gtan?/( d + Lsin?)] = v[9.81m/s²tan45.2° / (6.29m + 4.39msin45.2°)] = 1.025rad/s

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