A Ferris wheel with radius 9.5 m initially rotates at a constant rate by completing 0.75 rev in 25.00 s. It then begins to decelerate at a rate of 0.30 rad/s2 with its only passenger at the top of the wheel. Find the direction of the total acceleration vector. Express your answer as a positive number with respect to the positive x-axis. Choose up and to the right as your positive direction

Question

A Ferris wheel with radius 9.5 m initially rotates at a constant rate by completing 0.75 rev in 25.00 s. It then begins to decelerate at a rate of 0.30 rad/s2 with its only passenger at the top of the wheel. Find the direction of the total acceleration vector. Express your answer as a positive number with respect to the positive x-axis. Choose up and to the right as your positive direction

Answer 1

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Answer 2

To find the direction of the total acceleration vector, we need to consider both the centripetal acceleration and the tangential acceleration.

First, let's find the centripetal acceleration. The centripetal acceleration of an object moving in a circular path is given by the equation:

ac = ω^2 * r

where ac is the centripetal acceleration, ω is the angular velocity, and r is the radius of the circular path.

In this case, the angular velocity is given by:

ω = (2π * rev) / t

where t is the time taken to complete one revolution.

Substituting the given values, we have:

ω = (2π * 0.75) / 25.00 = 0.1508 rad/s

Now we can calculate the centripetal acceleration:

ac = (0.1508)^2 * 9.5 = 0.2148 m/s^2

Next, we need to find the tangential acceleration. The tangential acceleration is caused by the deceleration of the Ferris wheel.

The tangential acceleration is given by the equation:

at = α * r

where at is the tangential acceleration, α is the angular acceleration, and r is the radius of the circular path.

In this case, the angular acceleration is given by:

α = -0.30 rad/s^2 (negative because it is decelerating)

Substituting the radius, we have:

at = -0.30 * 9.5 = -2.85 m/s^2

Finally, we can find the total acceleration. The total acceleration vector is the vector sum of the centripetal acceleration and the tangential acceleration. Since the centripetal acceleration is always directed towards the center of the circle, and the tangential acceleration is in the opposite direction, the total acceleration will be the difference between the two.

|atotal| = |ac| - |at| = 0.2148 - 2.85 = -2.6352 m/s^2

Since we want the answer as a positive number, we take the absolute value:

atotal = |-2.6352| = 2.6352 m/s^2

Therefore, the magnitude of the total acceleration vector is 2.6352 m/s^2. The direction of the total acceleration vector is along the positive x-axis, as specified in the problem statement.