Speed in a Clockwise Circle

Mar 2012
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A particle starting from rest revolves with uniformly increasing speed in a clockwise circle in the xy plane. The center of the circle is at the origin of an xy coordinate system. At t = 0, the particle is at x = 0, y = 2 m. At t = 2 s, it has made one-quarter of a revolution and is at x = 2 m, y = 0. Determine (a) its speed at t = 2 s, (b) the average velocity vector, and (c) the average acceleration vector during this interval.

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.
 

HallsofIvy

MHF Helper
Apr 2005
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I would start by writing parametric equations for the circle: \(\displaystyle x= R cos(\theta)\), \(\displaystyle y= R sin(\theta)\), where R is the radius of the circle and the center is at (0, 0). It is easy to calculate that the velocity is given by \(\displaystyle -R\frac{d\theta}{dt} sin(\theta)\vec{i}+ R\frac{d\theta}{dt} cos(\theta)\vec{j}\) and that the acceleration is given by \(\displaystyle -R(\frac{d^2\theta}{dt} sin(\theta)+ (\frac{d\theta}{dt})^2 cos(\theta))\vec{i}+ R(\frac{d^2\theta}{dt^2}cos(\theta)- (\frac{d\theta}{dt})^2 sin(\theta))\vec{j}\).

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 \(\displaystyle \theta= (t^2+ \pi)/2\).
 
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Mar 2012
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If the speed is uniformly increasing, does it mean the acceleration is constant? If acceleration is constant why a = d^2θ / dt^2 one time and at = dθ / dt another time?
 
Jun 2008
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513
Illinois
If the speed is uniformly increasing, does it mean the acceleration is constant?
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 (\(\displaystyle a_r = \omega^2 R\)).

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.
 
Last edited:
Mar 2012
588
32
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?
 
Last edited:
Jun 2008
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Illinois
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. **
 
Last edited:
Mar 2012
588
32
It says at t = 2 s, it has made one-quarter of a revolution. Does that mean π/2 or π/4?

revolution = 2π

so (1/4)2π = π/2

What do you say?
 
Jun 2008
1,389
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Illinois
Yes, theta after two seconds is pi/2, my mistake. But that does not mean that velocity after 2 seconds is pi/4 --- that would be the average velocity of the first two seconds. Nor is alpha equal to pi/8. Use the formula I provided above to calculate alpha, with the correction that theta after 2 seconds is pi/2.
 
Mar 2012
588
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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

Now, what about the average velocity vector?

Pi m/2 s ????? What is its direction? northeast?
 
Last edited:
Jun 2008
1,389
513
Illinois
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
All good!

Now, what about the average velocity vector?

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