# A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t =

0

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

• 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

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