My willy bird is 6 inches and has a mass, M, of 2.1kg. During penetration, it reaches a force or 15.0 N with a coefficient of friction 0.75. Calculate its rotational velocity given no respect to time. How much work is done in a thirty second period?
To calculate the rotational velocity of the willy bird, we need to use the formula:
Angular velocity (ω) = Frictional force (F) / (Coefficient of friction (μ) * Mass (M) * Length (L))
Given:
Mass (M) = 2.1 kg
Force (F) = 15.0 N
Coefficient of friction (μ) = 0.75
Length (L) = 6 inches = 0.1524 meters (converting to meters)
First, let's convert the length from inches to meters:
Length (L) = 6 inches * 0.0254 meters/inch = 0.1524 meters
Now, we can substitute the given values into the formula:
Angular velocity (ω) = 15.0 N / (0.75 * 2.1 kg * 0.1524 m)
Simplifying the equation, we get:
Angular velocity (ω) = 15.0 N / (0.23418 kg*m)
Angular velocity (ω) = 64.01 rad/s (rounded to two decimal places)
So, the rotational velocity of the willy bird is approximately 64.01 rad/s.
Now, let's move on to calculating the work done in a thirty second period.
To calculate work (W), we use the formula:
Work (W) = Angular velocity (ω) * Torque (τ) * Time (t)
However, we are not given the torque (τ) in this case. Without knowledge of the torque, it is not possible to determine the work done.