what is its angular acceleration in rad/s2 ? Answer in units of rad/s 2 .

Question

A potter’s wheel moves from rest to an angular speed of 0.30 rev/s in 28.9 s. Assuming constant angular acceleration, what is its angular acceleration in rad/s2 ? Answer in units of rad/s 2 .

Answer 1

ε = ω/t = 0.30rev/s/28.9s = 0.01038 rev/s² = 2π·0.01038 rad/s² = 0.0652 rad/s²

Answer 2

From the question given, we can see that it gives us three known quantities.

Initial rotation speed ω₀: 0 rev/s (the "wheel moves from rest")

Final rotation speed ω_f: 0.30 rev/s

The time over which the change occurs t: 28.9s

The question also tells us that the angular acceleration, the unknown quantity we're looking for, is constant. This means the velocity is changing at the same rate across the 28.9s where the wheel is spinning. So, we can use an equation that assumes a constant angular acceleration α.

α = (ω_f - ω₀)/t

Now, here one must note that the question asks for the answer in different units than the quantities given. Thus, once one calculates α, one must convert it to rad/s². Since 1 revolution = 2π radians:

α (in rev/s²) * (2π radians/1 revolution) = α (in rad/s²)

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Tangent explanation of why the equation for α has the form it does:

This equation has the same form as a similar problem involving linear motion, just with the velocity and acceleration variables swapped out for their angular counterparts. Here's how the equation gives you angular acceleration (the rate of change of the angular velocity): it takes the whole change in velocity across the whole time period in question and splits it into chunks such that α is how much velocity the wheel gains (or loses if it's slowing down) in one second. α's units of rev/s^2 = rev/s/s corresponds to this. The assumption of constant acceleration is required because it allows us to assume the velocity gained (or lost) each chunk will be the same. If acceleration wasn't constant, the wheel could be gaining (or losing) a different amount of velocity in each one-second chunk, and this equation would glaze over such detail. Hopefully this explanation will help you reason the correct form of the equation rather than having to memorize it for tests :)

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