A particular Ferris wheel (a rigid wheel rotating in a vertical plane about a horizontal axis) at a local carnival has a radius of 20.0 m and it completes 1 revolution in 9.84 seconds.

(a) What is the speed (m/s) of a point on the edge of the wheel?
Using the coordinate system shown, �nd:
the (b) x component of the acceleration of point A at the top of
the wheel.
the (c) y component of the acceleration of point A at the top of the wheel
the (d) x component of the acceleration of point B at the bottom of the wheel.
the (e) y component of the acceleration of point B at the bottom of the wheel
the (f) x component of the acceleration of point C on the edge of the wheel.
the (g) y component of the acceleration of point C on the edge of the wheel

(a) The speed of a point on the edge of the wheel can be calculated using the formula:

speed = 2πr / time

In this case, the radius (r) is 20.0 m and the time is 9.84 seconds. Plugging in these values:

speed = (2 * π * 20.0) / 9.84 ≈ 12.8 m/s

(b) At the top of the wheel, point A is moving horizontally. Therefore, the x-component of its acceleration is zero.

(c) At the top of the wheel, point A is moving vertically downwards, so the y-component of its acceleration is directed downwards.

(d) At the bottom of the wheel, point B is moving horizontally. Therefore, the x-component of its acceleration is zero.

(e) At the bottom of the wheel, point B is moving vertically upwards, so the y-component of its acceleration is directed upwards.

(f) At point C on the edge of the wheel, the x-component of its acceleration is zero since it is moving vertically.

(g) At point C on the edge of the wheel, the y-component of its acceleration is constant, directed towards the center of the wheel.

To solve this problem, we need to use the following equations:

(a) Speed (v) = Circumference / Time period
(b) x-component of acceleration (a_x) = -v^2 / r
(c) y-component of acceleration (a_y) = g
(d) x-component of acceleration (a_x) = -v^2 / r
(e) y-component of acceleration (a_y) = -g
(f) x-component of acceleration (a_x) = -v^2 / r
(g) y-component of acceleration (a_y) = -g

Given:
Radius (r) = 20.0 m
Time period (T) = 9.84 seconds
Gravitational acceleration (g) = 9.8 m/s^2

Let's solve each part step-by-step:

(a) Speed (v) = Circumference / Time period
The circumference of a circle is given by 2 * π * r.
Substitute the values: v = (2 * π * r) / T

(b) x-component of acceleration (a_x) = -v^2 / r
Substitute the value of v from the previous step: a_x = -(v^2 / r)

(c) y-component of acceleration (a_y) = g
Given as g = 9.8 m/s^2

(d) x-component of acceleration (a_x) = -v^2 / r
Substitute the value of v from step (a): a_x = -(v^2 / r)

(e) y-component of acceleration (a_y) = -g
Given as g = 9.8 m/s^2

(f) x-component of acceleration (a_x) = -v^2 / r
Substitute the value of v from step (a): a_x = -(v^2 / r)

(g) y-component of acceleration (a_y) = -g
Given as g = 9.8 m/s^2

Let's calculate each step one by one.

(a) Speed:
v = (2 * π * r) / T
= (2 * 3.1415 * 20.0) / 9.84
≈ 12.76 m/s

(b) x-component of acceleration at point A:
a_x = -(v^2 / r)
= -(12.76^2 / 20.0)
≈ -8.12 m/s^2

(c) y-component of acceleration at point A:
a_y = g = 9.8 m/s^2

(d) x-component of acceleration at point B:
a_x = -(v^2 / r)
= -(12.76^2 / 20.0)
≈ -8.12 m/s^2

(e) y-component of acceleration at point B:
a_y = -g = -9.8 m/s^2

(f) x-component of acceleration at point C:
a_x = -(v^2 / r)
= -(12.76^2 / 20.0)
≈ -8.12 m/s^2

(g) y-component of acceleration at point C:
a_y = -g = -9.8 m/s^2

So, the answers are:
(a) Speed = 12.76 m/s
(b) x-component of acceleration at point A = -8.12 m/s^2
(c) y-component of acceleration at point A = 9.8 m/s^2
(d) x-component of acceleration at point B = -8.12 m/s^2
(e) y-component of acceleration at point B = -9.8 m/s^2
(f) x-component of acceleration at point C = -8.12 m/s^2
(g) y-component of acceleration at point C = -9.8 m/s^2

To find the speed of a point on the edge of the Ferris wheel, you can use the formula for linear speed:

Speed = Circumference / Time

The circumference of a circle can be found using the formula:

Circumference = 2 * π * radius

Given that the radius of the Ferris wheel is 20.0 m and it completes 1 revolution in 9.84 seconds, we can substitute these values into the formula:

Circumference = 2 * π * 20.0 m = 125.6 m

Now, we can calculate the speed:

Speed = Circumference / Time = 125.6 m / 9.84 s ≈ 12.76 m/s

So, the speed of a point on the edge of the Ferris wheel is approximately 12.76 m/s.

To find the x component of the acceleration at point A (top of the wheel), we need to identify the forces acting on that point. At the top of the wheel, we have the weight force acting straight down and the tension force acting horizontally. The x component of the acceleration can be found using Newton's second law:

Force x = mass * acceleration x

Since the point is at rest horizontally, the net force in the x direction is zero. Therefore, the x component of the acceleration at point A is zero.

To find the y component of the acceleration at point A (top of the wheel), we can use circular motion equations. The acceleration of an object moving in a circle can be found using the formula:

Acceleration = (velocity^2) / radius

At the top of the Ferris wheel, the velocity is zero at that instant, so the y component of the acceleration at point A is zero.

Similarly, for point B (bottom of the wheel), the x component of the acceleration is zero because the wheel is at rest horizontally.

For the y component of the acceleration at point B, we can again use the circular motion equation:

Acceleration = (velocity^2) / radius

At the bottom of the Ferris wheel, the velocity is at its maximum, which is equal to the speed we calculated earlier (12.76 m/s). Since the radius is also given as 20.0 m, we can calculate the y component of the acceleration using the formula:

Acceleration = (12.76 m/s)^2 / 20.0 m = 8.16 m/s^2

Moving on to point C (edge of the wheel), the x component of the acceleration is still zero because the wheel is at rest horizontally.

For the y component of the acceleration at point C, we can use the same circular motion equation:

Acceleration = (velocity^2) / radius

Since point C is moving with the same speed as the point on the edge of the wheel, the y component of the acceleration at point C is the same:

Acceleration = (12.76 m/s)^2 / 20.0 m = 8.16 m/s^2

To summarize the answers:
(a) The speed of a point on the edge of the wheel is approximately 12.76 m/s.
(b) The x component of the acceleration at point A is zero.
(c) The y component of the acceleration at point A is zero.
(d) The x component of the acceleration at point B is zero.
(e) The y component of the acceleration at point B is 8.16 m/s^2.
(f) The x component of the acceleration at point C is zero.
(g) The y component of the acceleration at point C is also 8.16 m/s^2.

(a) Divide 2*pi*R by 9.84 s for the speed.

The speed is suspiciously high, and probably unsafe.

(b) and (c) At the top of the wheel, the centripetal acceleration is down (-y). Its magnitude is V^2/R everywhere along the outer edge

(d) and (e) At the bottom of the wheel, the centripetal acceleration is up (+y)

(f) and (g) It depends upon which edge. Location C is not shown.