| Problem statement Solution video |
DISCUSSION THREAD
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DISCUSSION and HINTS
As expected, the acceleration of P has both non-zero tangential and normal components.
- From the equation provided for speed as a function of distance traveled, the speed of P is monotonically increasing over time. Therefore, the tangential component of acceleration always points "forward" of the direction of motion.
- Although the speed of P is increasing, the normal component appears to decrease as P moves along its path - why is that?
Can you see these two things in the animation below?
Recall that the general path description velocity and acceleration equations are given by the following:
v = v*et
a = v_dot*et + (v2/ρ)*en
Part (b)
Note that v_dot = dv/dt. For this problem, we do NOT know the speed as a function of time; instead, we know speed as a function of position, s. To find v_dot, we need to use the chain rule of differentiation: dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds).
Part (c)
Use your sketch in Part (a) to write et and en in terms of i and j. Substitute these in your results from Part (b) to find the x- and y-components of velocity and acceleration.
Part (d)
The center of curvature of the path, C, is located in the en direction at a distance of ρ from P.
For part d, is the answer expected as a point or in vector form? The hint also mentions the e-sub-n direction but I feel finding the values using trigonometry works too. Will we be deducted for using a different than expected process?
Part d) is asking for the xy-components of the location of the center of curvature. You can just give those components, or you may write the position of that point measured relative to the origin as a vector.
Hey Nathan, the way I see it, using trigonometry or the e-sub-n direction "method" is actually two different ways of saying the same thing.
The way I approached this problem is by converting e-sub-n-hat to i and j (x and y) components. Because we know that e-sub-n-hat is a unit vector, it has components of sin(theta)i - cos(theta)j. Knowing this form of the unit vector, we can multiply this by the scalar value rho in order to find the directional vector for the location of C.
This method also uses trigonometry, just in the conversion to Cartesian coordinates.
For part c determining the i and j components, the angle you would use for en is the angle between the y-axis and the en vector right? And that would be equivalent to the angle between the x-axis and et?
Disregard the question, I made a mistake
for part a for solving for the dv/ds, would you get 2bs * b s dot ^squared? I'm a little confused on how to solve this derivative
Ella:
It is given that: v = b*s^2
From this, we have (using the chain rule of differentiation):
dv/dt = (dv/ds)*(ds/dt) = (2*b*s)*s_dot = (2*b*s)*v
Let us know if this does not make sense to you.
I assume you are meaning to say part b, but to solve part b you want to do dv/dt and then use chain rule to do dv/ds * (ds/dt). Where ds/dt is equal to velocity.
velocity is always tangent to the path and we can find acceleration using the chain rule
When turning CCW, is the normal component of acceleration supposed to be negative? If so, why?
Sorry, when turning CW should it be negative?
The direction of the normal unit vector e_n is set by two things. One, it is PERPENDICULAR to the path. Two, it points INWARD to the path. Since the component of acceleration in the e_n direction is always positive (v^2/rho > 0), the normal component of acceleration always points in the positive e_n direction.
This is true regardless of the direction in which the path turns.
For this problem, e_n points from P toward C.
Would it be correct to think of it as de_t/dtheta is positive when rotating CCW as theta is increasing relative to the positive x-axis? And that e_n should be multiplied by -1 when rotating CW to account for theta decreasing relative to the positive x-axis?
Try not to over-complicate the problem. Here is what we know about the problem:
* We are given speed, v= b*s^2. That is the only component of velocity. No need for additional work needed to find velocity.
* To find the tangential component of acceleration, you need to find v_dot = dv/dt = (dv/ds)(ds/dt). See above discussion. To find the normal component of acceleration, use v^2/rho. You have expressions for both of those. This completes Part (b).
* To find the Cartesian components, write e_t and e_n in terms of i and j using trig. Substitute these into your expressions for velocity and acceleration to find the Cartesian components.
When the question asks for all the values at P where s=3, do we assume that is the point on the shown graph? I assume so, as otherwise the rho and theta values would be useless, but it doesn't say it explicitly.
Yes, all values are to be found at the same value of s = 3, the position shown in the figure.