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Hatshy [7]
3 years ago
12

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo

w has a 5.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?
Physics
1 answer:
Valentin [98]3 years ago
3 0

We know that the acceleration due to gravity g is: g = 9.81 m/s^2

So the centripetal acceleration (w) is:

w^2 = 1.5 g / r

w^2 = 1.5 * (9.81 m/s^2) / 5 m

w = 1.716 rad / s

To convert to rad to rev:

w = (1.716 rad / s) * (1 rev / 2π rad) * (60 s/min)

<span>w = 16.4 rev/min </span>

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The chance in distance is 25 knots

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s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

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\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

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\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

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