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Set 8 Problem number 7
A velocity of 57.99 mph, according to the
approximation used here, is ( 57.99 / 2.3) m/s = 25.21 m/s.
- At this speed, on a circle of radius 125 meters, the
centripetal acceleration, or the acceleration toward the center, would be
- centripetal acceleration = v^2 / r = ( 25.21 m/s) ^
2 / 125 meters = 5.085 meters per second per second.
1 mile / hour = 5280 feet / 3600 sec = 5280 * .305
meters / 3600 sec = 1600 m / 3600 sec = .44 m/s, so 1 m/s = (1 / .44 ) mph = 2.25 mph,
which we will approximate as 2.3 mph.
A velocity of 57.99 mph, according to this
approximation, is ( 57.99 / 2.3) m/s = 25.21 m/s.
- At this speed, on a circle of radius 125 meters, the
centripetal acceleration (the acceleration toward the center), would be
- centripetal acceleration = v^2 / r = ( 25.21 m/s) ^
2 / 125 meters = 5.085 meters per second per second.
- The force required to accelerate a 1574 kg mass at
this rate is F = ma = ( 1574 kg)( 5.085 m/s/s) = 8003 Newtons.
- The centripetal acceleration of an object moving
along a circle of radius r at velocity v is a = v^2 / r; this acceleration is directed
toward the center of the circle.
- If the object has mass m, the force required to keep
the object in its circular path is F = m a = m v^2 / r; this force is also directed toward
the center of the circle.
Explanation in terms of Figure(s); Extension
The figure below shows an object moving with
constant speed v on a circle of radius r.
- The acceleration required to keep the object moving
in the circle, as opposed to the straight line along which it would travel if it had no
acceleration, is a = v^2 / r, and is directed toward the center of the circle.
- If the object has mass m, the force required to keep
it on its circular path will therefore be F = m a = m v^2 / r. This force will be directed
in the same direction as the acceleration, toward the center of the circle.
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