In the last problem, assume the tangential acceleration is constant and determine the components of the instantaneous acceleration at (a) t = 0, (b) t = 1 s, and (c) t = 2 s.

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In the last problem, assume the tangential acceleration is constant and determine the components of the instantaneous acceleration at (a) t = 0, (b) t = 1 s, and (c) t = 2 s.

Since the speed is uniformly increasing, we must have that \(\displaystyle \frac{d^2\theta}{dt^2}= a\), a constant, so that \(\displaystyle \frac{d\theta}{dt}= at\) and \(\displaystyle \theta= (t^2+ \pi)/2\).

No - not for a particle moving in a circle. Recall that acceleration for a particle in constant circular motion (where angular acceleration = 0) is a vector of magnitude \(\displaystyle \omega^2 R\). where omega = rotation rate in rad/s, and the direction of that vector points toward the center of rotation. In this problem the angular acceleration is not constant, so the total acceleration of the particle is the vector sum of its angular acceleration in a direction tangential to the circle (\(\displaystyle a_t= \alpha R\) ) plus the radial acceleration towards the center (\(\displaystyle a_r = \omega^2 R\)).If the speed is uniformly increasing, does it mean the acceleration is constant?

To solve this I would start by determining the value for alpha. You can use \(\displaystyle \theta = \frac 1 2 \alpha t^2\) to determine alpha from the data provided. From that you can then determine the speed at t= 2 seconds from \(\displaystyle \omega = \alpha t\), and speed = \(\displaystyle \omega R\). I'm not entirely sure what they mean by the "average velocity vector," although I suspect what they want is the velocity vector at the instant that the velocity is half its average magnitude for the first two seconds - i.e. the velocity vector at t= 1 second. And I would do the same for the "average acceleration vector" - the vector sum of radial plus tangential acceleration at t= 1 second.

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α = dω / dt

ω = (1/4)rev/2 s = π rad/4 s

Can I say that α = Δω / Δt = (π rad/4 s) / 2 s = π rad/8 s^2?

speed = Rω = (2 m)(π rad/4 s) = π m/2 s = 1.57 m/s

average velocity vector = Δx / Δt = [ (2 m)i - (2 m)j ] / 2 s = (i - j) m/s

average acceleration vector = Δv / Δt = (i - j) m/s / 2s = [ (0.5)i - (0.5)j ] m/s^2

Did I make anything wrong?

ω = (1/4)rev/2 s = π rad/4 s

Can I say that α = Δω / Δt = (π rad/4 s) / 2 s = π rad/8 s^2?

speed = Rω = (2 m)(π rad/4 s) = π m/2 s = 1.57 m/s

average velocity vector = Δx / Δt = [ (2 m)i - (2 m)j ] / 2 s = (i - j) m/s

average acceleration vector = Δv / Δt = (i - j) m/s / 2s = [ (0.5)i - (0.5)j ] m/s^2

Did I make anything wrong?

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No, that won't work - you are assuming that \(\displaystyle \omega\) is constant, but it's not. Start with:

\(\displaystyle \theta = \frac 1 2 \alpha t^2\)

At t = 2 s you know that \(\displaystyle \theta = \pi/4\), so from that what do you get for \(\displaystyle \alpha\)?

**EDIT - that should be \(\displaystyle \theta = \pi/2\), as noted in subsequent posts. **

\(\displaystyle \theta = \frac 1 2 \alpha t^2\)

At t = 2 s you know that \(\displaystyle \theta = \pi/4\), so from that what do you get for \(\displaystyle \alpha\)?

**EDIT - that should be \(\displaystyle \theta = \pi/2\), as noted in subsequent posts. **

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All good!Pi / 2 = (1/2)alpha(4 s^2) =>

Pi / 2 = alpha(2 s^2) =>

Pi / 4 s^2 = alpha

ω = alpha(t) = (Pi / 4 s^2)(2 s) = Pi / 2 s

speed = Rω = (2 m)( Pi / 2 s) = Pi m/s

Pi/2 m/s is what I would use for the magnitude. But given that the particle is moving clockwise starting from (0,2), it's direction at t=1 sec would not be toward the northeast. Draw a figure and you'll see.Now, what about the average velocity vector?

Pi m/2 s ????? What is its direction? northeast?

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