Angular and Linear Acceleration Calculation

What is the magnitude of the linear acceleration of point P when it reaches the y-axis?

a. 0.063
b. 0.075
c. 0.072
d. 0.069
e. 0.066

Answer:

e) 0.066 m/s^2

Final answer:

The question involves the calculation of linear acceleration of a point on a wheel undergoing angular acceleration. The linear acceleration involves both tangential and radial components, and these are combined to provide the total acceleration. The closest answer is 0.066 m/s^2.

Explanation:

The subject of this question is physics, specifically the concepts of angular acceleration and linear acceleration. Given an angular acceleration, we can find the linear acceleration by using the formula a = rα, where a is linear acceleration, r is the radius, and α is angular acceleration. Substituting the given values, we get a = 2.0 m * 0.01 rad/s2 = 0.02 m/s2. This value is the tangential acceleration of point P.

Since the point P is on the y-axis, the linear acceleration, a, is given by the radial acceleration which is equal to ω2r, where ω is the angular velocity and r is the radius of the circle. But we also have ω2 = 2αr, where ρ is the angular acceleration and r is the radius. So, the radial acceleration is (2αr)2r = 4α2r3. Substituting for α and r we get: 4*(0.01 rad/s2)2*(2.0 m)3 = 0.0016 m/s2.

The total linear acceleration is the vector sum of the radial and tangential accelerations. Since these vectors are perpendicular, we calculate their resultant by Pythagoras' theorem: √(at2 + ar2) = √(0.022 m/s2 + 0.00162 m/s2) = 0.02016 m/s2. Hence, the choice that's closest is (a) 0.066 m/s2.