Rotational kinematics and cen ral College-PHYS 1111-Fall17-SEAT Activities and D
ID: 1782558 • Letter: R
Question
Rotational kinematics and cen ral College-PHYS 1111-Fall17-SEAT Activities and Due Dates Gr /2017 11:00 PM 62/100 11/14/2017 09:57 PM Print CalcuiatorPeriodic Table 1 of 29 pling Learning A speedboat of mass 545 kg (including the driver) is tethered to a fixed buoy by a strong 30.7-m cable. The boat's owner loves high speed, but doesn't really want to go anywhere. So he revs up the boat's engine, makes a lot of noise, and runs the boat in circles around the buoy with the cable supplying all the necessary centripetal force. When the tension of the cable is steady at 1.33 x 10 N, with what force is the boars engine pushing the boat? Different physics textbooks treat drag force somewhat differently and use different formulas. For the present purpose, take the water's drag force on the boat to be (450 kg/m)xv, where v denotes the boat's speed. Ignore any drag force on the cable. Numbar check Answer NextExplanation / Answer
Well, buoys are anchored to the seabed by a chain with enough slack to allow for tidal changes plus worst-case sea-state. I.e. they move. Let's assume instead that this idiot has hooked themselves to a ring around a piling that lets them go around without being wound into it like a yo-yo. And assume that somehow the cable is held above the water and has no drag, yet despite being attached high up on the speedboat well above its center of mass, does not pull it over sideways and make it capsize. No, forget the cable, we'll install a rigid vertical massless post in the boat welded to a rigid massless bar with the other end welded to a sleeve bearing on another post driven solidly into the seabed.
Yes, Fc=(mv^2)/r will give you the velocity since you know m and r and Fc. So then you can get the drag from 450v^2
So, we proceed:
v^2 = Fc*r/m = (1.33*104)(30.7)/(545) = 749.19
So,
Drag = 450*v^2
= 450 * 749.19
= 3.37 x 105 N
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