A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?
We have to first determine the normal and tangential components of acceleration to determine the total acceleration of the point on the edge of the wheel.
The radius of the wheel, r = 20 cm / 2 = 10 cm = 0.1 m
The tangential acceleration of the point, (i.e. acceleration of the point tangential to, and in the direction of, its path of motion),
t = α * r,
where α is the angular acceleration of the wheel, and r is the radius of the wheel.
=> t = (2.0 rad/s²) * (0.1 m) = 0.2 m/s²
The normal acceleration of the point, (i.e. the acceleration of the point normal to the path of motion, and pointing in towards the radius of curvature of the path),
n = ω² * r,
where ω is the angular velocity of the wheel (at t = 0.60 s), and r is the radius of the wheel.
To determine ω at t = 0.60s, let’s turn to kinematics:
α = ω / t
=> ω = α * t = (2.0 rad/s²) * (0.60 s) = 1.2 rad/s
Now we can determine the normal acceleration of the point:
n = ω² * r = (1.2 rad/s)² * (0.1 m) = 0.144 m/s²
The total acceleration of the point on the outer edge of the wheel,
a = √(t² + n²) = √[(0.2 m/s²)² + (0.144 m/s²)²] = 0.246 m/s²
Therefore, the total acceleration of the point on the outer edge of the wheel at t = 0.60s is: 0.25 m/s².Source(s): Hibbeler, R.C. Engineering Mechanics – Dynamics. Third Edition. Pearson Prentice Hall: 2004.
3. (B). They will have the same velocity v = √(2gh/(1+k)) where k = ½ (I = kmr²) Ergo, the angular velocity wl = v/r will be ½ws, since its r value is twice that of ws