A Ferris wheel with radius 14.0m is tuurning about a horizontal axis through its center. The linear speed of a passenger on the rin is constant and equal to 7.00 m/s. Whatare the magnitude and direction of the passengers acclerations as she passes through a) the lowest point in her circular motion and b) the highest point in her circular motion? c) how much time does it take the ferris wheel to make on revolution?
(a), (b)
acceleration due to circular motion
= rω2 towards the centre
It is necessary to add/subtract the acceleration due to gravity at different points on the wheel
(c)
tangential speed, v= rω
v and r are know, so ω can be determined.
For C i got .5 or 1/2 v=rw
7.0=14.0 w is that correct
im still a little confused on a and b
at the top, net acceleration=centacc-9.8
at the bottom, add them to get net acceleration
centripetal acceleration=v^2/r r, v given.
distanceAround/time=v
solve for time.
That's correct. The unit is radians / s.
For a and b, the acceleration I was referring to corresponds to the centripetal force required to keep the passenger in place.
Thus at the lowest point, the weight of the passenger adds to this centripetal force, and so the acceleration due to gravity, g, is additve to this acceleration.
On the other hand, at the highest point, the acceleration due to gravity tends to reduce the centripetal force required to keep the passenger in its place, so g must be subtracted from the centripetal acceleration that you calculate from the formula a=ω2r.
To find the magnitude and direction of the passenger's acceleration at different points on the Ferris wheel, we can use the following formulas:
a) At the lowest point:
The linear speed of the passenger and the radius of the Ferris wheel can be used to find the tangential acceleration (at) at that point, which is directed tangentially to the circular path.
at = v^2 / r
Plugging in the values:
v = 7.00 m/s
r = 14.0 m
at = (7.00 m/s)^2 / 14.0 m
at = 49.00 m^2/s^2 / 14.0 m
at ≈ 3.50 m/s^2
The magnitude of the acceleration at the lowest point is approximately 3.50 m/s^2, directed tangentially to the circular motion.
b) At the highest point:
The linear speed of the passenger and the radius of the Ferris wheel can be used to find the centripetal acceleration (ac) at that point, which is directed towards the center of the circular path.
ac = v^2 / r
Plugging in the values:
v = 7.00 m/s
r = 14.0 m
ac = (7.00 m/s)^2 / 14.0 m
ac = 49.00 m^2/s^2 / 14.0 m
ac ≈ 3.50 m/s^2
The magnitude of the acceleration at the highest point is also approximately 3.50 m/s^2, directed towards the center of the circular path.
c) To find the time it takes for the Ferris wheel to make one revolution, we can use the formula for the period (T) of an object in uniform circular motion:
T = 2πr / v
Plugging in the values:
r = 14.0 m
v = 7.00 m/s
T = 2π(14.0 m) / 7.00 m/s
T = 2π(2.00 s)
T ≈ 12.57 s
So, it takes approximately 12.57 seconds for the Ferris wheel to make one revolution.