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
To find the direction of the total acceleration vector, we need to consider the two components of acceleration: tangential acceleration (a_t) and centripetal acceleration (a_c).
First, let's find the tangential acceleration. Tangential acceleration is the rate at which the speed of an object is changing. It can be calculated using the formula:
a_t = (Δv) / Δt
where:
Δv is the change in velocity
Δt is the change in time
Since the Ferris wheel is initially rotating at a constant rate, there is no change in velocity, and therefore no tangential acceleration. Thus, a_t = 0.
Next, let's find the centripetal acceleration. Centripetal acceleration is the acceleration towards the center of a circular path and can be calculated using the formula:
a_c = (v^2) / r
where:
v is the velocity
r is the radius of the circular path
To find the velocity, we know that the Ferris wheel completes 0.75 revolutions in 25.00 seconds. We can find the angular velocity (ω) using the formula:
ω = (Δθ) / Δt
where:
Δθ is the change in angle (in radians)
Δt is the change in time
Since the Ferris wheel completes 0.75 revolutions (0.75 * 2π radians) in 25.00 seconds, we have:
ω = (0.75 * 2π rad) / 25.00 s
Now, using the formula for velocity in circular motion:
v = ω * r
We can substitute the value of ω and the radius of the Ferris wheel to find the velocity.
Finally, we can substitute the values for the velocity and the radius in the formula for centripetal acceleration to find the magnitude of the centripetal acceleration, a_c.
Now that we have both the tangential acceleration (a_t = 0) and the centripetal acceleration (a_c), we can find the direction of the total acceleration vector by considering their relative magnitudes.
Since the only passenger is at the top of the wheel, the total acceleration vector will point towards the center of the Ferris wheel. Therefore, the direction of the total acceleration vector is downward, towards the positive y-axis, as given in the problem statement.