A fan blade is rotating with a constant angular acceleration of +12.4 rad/s2. At what point on the blade, as measured from the axis of rotation, does the magnitude of the tangential acceleration equal that of the acceleration due to gravity?
I do not know how to start this.
ε= 12.4 rad/s²,
a(tang) =9.8 m/s²
a(tang) = ε•R.
R = a(tang)/ ε = 9.8/12.4 = 0.79 m
To solve this problem, we need to find the distance from the axis of rotation at which the magnitude of the tangential acceleration equals that of the acceleration due to gravity. Let's break down the problem step by step:
1. Recall that the tangential acceleration can be calculated using the equation:
a_tangential = r * α
where a_tangential is the tangential acceleration, r is the distance from the axis of rotation, and α is the angular acceleration.
2. The acceleration due to gravity, g, is a constant value of approximately 9.8 m/s^2.
3. We are given that the angular acceleration is +12.4 rad/s^2. Note that the "+" sign indicates that the rotation is in the counter-clockwise direction.
4. We want to find the value of r when the magnitudes of the tangential acceleration and acceleration due to gravity are equal, or in other words, when a_tangential = g.
Now, let's solve for r:
a_tangential = g
r * α = g
r = g / α
Substituting the given values, we have:
r = 9.8 m/s^2 / 12.4 rad/s^2
By dividing 9.8 m/s^2 by 12.4 rad/s^2, we can find the value of r in meters.
To solve this problem, we need to compare the magnitudes of the tangential acceleration and the acceleration due to gravity at a certain point on the fan blade.
First, let's write down the given values and the known formulas:
Angular acceleration (α) = +12.4 rad/s²
Acceleration due to gravity (g) = 9.8 m/s²
Radius of the fan blade (r) = ?
We know that the tangential acceleration (at) can be calculated using the formula at = r * α, where r is the distance from the rotation axis to the point on the blade.
Acceleration due to gravity is given by the formula ag = g.
We want to find the point on the blade where the magnitudes of at and ag are equal. So we can set up the equation:
at = ag
r * α = g
Now, we can solve for the radius (r):
r = g / α
Substituting the values, we get:
r = 9.8 m/s² / 12.4 rad/s²
Simplifying, we find:
r ≈ 0.79 m
Therefore, at a distance of approximately 0.79 meters from the axis of rotation, the magnitude of the tangential acceleration will be equal to the acceleration due to gravity.