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The original Ferris wheel was built in 1893 by George Washington Gale Ferris, Jr., a civil engineering. The wheel, an amazing engineering construction at the time, carried 36 wooden cars, each holding as many as 60 passengers, around a circle of radius R = 38 m. The mass of each car was about 1.1xx10^(4)kg. The mass of the wheel's structure was about 6.0xx10^(5)kg, which was mostly in the circular grid from which the cars were suspended. The wheel made a complete rotation at an angular speed omega_(F) in about 2 min. (a) Estimate the magnitude L of the angular momentum of the wheel and its passengers while the wheel rotated at omega_(F). (b) If the fully loaded wheel is rotated from rest to omega_(F) in a time period Deltat_(1)=5.0s, what is the magnitude t_(avg) of the avergae net external torque acting on it? |
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Answer» Solution :(a) We can treat the wheel, cars, and passengers as a rigid object rotating about a fixed axis, at the wheel.s axle. Then Eq. `(L=Iomega)` gives the magnitude of the angular momentum of that object. We NEED to find `omega_(F)` and the rotational inertia I of this object. Calculations: Rotational inertia: To find I, let us start with the loaded cars. Because we can treat them as particles, at distance R from the axis of rotation, we know from Eq. That their rotational inertia is `I_(pc)=M_(pc)R^(2),"where "M_(pc)` is their total mass. Let us assume that the 36 cars are each filled with 60 passengers, each of mass 70 kg. Then their total mass is `M_(pc)=36[1.1xx10^(4)kg+60(70kg)]=5.47xx10^(5)kg` and their rotational inertia is `I_(pc)=M_(pc)R^(2)=(5.47xx10^(5)kg)(38m)^(2)=7.90xx10^(8)kg*m^(2)` Next, we consider the structure of the wheel. Let us assume that the rotational inertia of the structure is due mainly to the circular grid suspending the cars. Further, let us assume that the grid forms a hoop of radius R, with a mass `M_("hoop")" of "3.0xx10^(5)kg` From, the rotational inertia of the hoop is `I_("hoop")=M_("hoop")R^(2)=(3.0xx10^(5)kg)(38m^(2))` = `4.33xx10^(8)kg*m^(2)` The combined rotational inertia I of the cars, passengers, and hoop is then `I=I_(pc)+I_("hoop")=7.90xx10^(8)kg*m^(2)+4.33xx10^(8)kg*m^(2)` = `1.22xx10^(9)kg*m^(2)` Angular speed : TO find the rotational speed `omega_(F)`, we use `omega_(avg)=Deltatheta//DELTAT`. Here the wheel goes through an angular DISPLACEMENT of `Deltatheta=2pi` rad in a time period `Deltat=2min`. Thus, we have `omega_(F)=(2pirad)/((2min)(60s//min))=0.0524rad//s` Angular momentum: Now we can find the magnitude L of the angular momentum with Eq. `L=Iomega_(F)=(1.22xx10^(9)kg*m^(2))(0.0524rad//s)` = `6.39xx10^(7)kg*m^(2)//s~~6.4xx10^(7)kg*m^(2)//s` (b) Conceptualize/Classify: The average NET external torque is related to the change `DeltaL` in the angular momentum of the loaded wheel by Eq. `(vectau_("net")=dvecL//dt)`. Compute: Because the wheel rotates about a fixed axis to reach angular speed `omega_(F)` in time period `Deltat_(1)`, we can reqriye Eq. `tau_(avg)=DeltaL//Deltat_(1)` The change `DeltaL` is from zero to the answer for part (a). Thus, we have `tau_(avg)=(DeltaL)/(Deltat_(1))=(6.39xx10^(7)kg*m^(2)//s-0)/(5.0s)` `~~1.3xx10^(7)N*m` |
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