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?
0.25 m/s2
0.50 m/s2
0.14 m/s2
0.34 m/s2
0.20 m/s2
2 Answers

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