Question

A motorized wheel is spinning with a rotational velocity of at the operator switches the wheel to a higher speed setting. the rotational velocity of the wheel at all subsequent times is given by where and 0 = 1.57 rad/s. = 0 s, () = 0 arctan (c ) 2 c = 1.5574 = 5.87 s . −2 at what time max is the rotational acceleration at a maximum? max = ms what is the maximum tangential acceleration of a point on the wheel a distance = 1.19 m from its center? = m/s 2 what is the magnitude of the total acceleration of this point?

Answer 1

The maximum tangential acceleration of a point on the wheel is 0.714 m/s^2. The magnitude of the total acceleration of this point is 425.34 m/s^2.

Step-by-step explanation:

To determine the maximum tangential acceleration of a point on the wheel, we need to calculate the angular acceleration first. The angular acceleration can be calculated using the formula:

α = (Δω) / (Δt)

We have Δω = 3 rev/s and Δt = 5 s, so α = (3 rev/s) / (5 s) = 0.6 rev/s^2.

Next, we can calculate the tangential acceleration using the formula:

at = rα

Given the distance from the center of the wheel, r = 1.19 m, we can calculate:

at = (1.19 m)(0.6 rev/s^2) = 0.714 m/s^2.

The magnitude of the total acceleration of this point can be calculated using the formula:

a = √(at^2 + ar^2)

Since the wheel is rotating in a circle, the radial acceleration, ar, is given by:

ar = ω^2r

where ω is the angular velocity in rad/s. At t = 5.87 s, ω = 3 rev/s * (2π rad/rev) = 18.85 rad/s. Plugging in the values:

ar = (18.85 rad/s)^2 * (1.19 m) = 425.34 m/s^2.

Now we can calculate the total acceleration:

a = √((0.714 m/s^2)^2 + (425.34 m/s^2)^2) = 425.34 m/s^2.