Homework H1.C - Sp23

[/embed]

Problem statement
Solution video

DISCUSSION THREAD

Ask and answer questions here. You learn both ways.


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 decreasing over the range of motion shown in the animation below. Therefore, the tangential component of acceleration always points "backward" of the direction of motion.
  • The normal component decreases as P moves along the track since the speed of P is decreasing.

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

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).

34 thoughts on “Homework H1.C - Sp23”

    1. Yes, v_dot is found from the differentiation of v with respect to TIME, "t". Note that v is given as a function of "s". To find dv/dt, you need to use the chain rule: dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds).

  1. After doing the differentiation mentioned above I believe we would get an expression for v_dot which we plug in the s value at positon b. I was wondering if it is correct to say this expression is also an expression in terms of s.

  2. When trying to find the values at point B, do we say that "s" in cos(bs) is 1/4 of the circumference since that will cancel with the units of b? Or is that not allowed because that isn't technically an angle?

    1. Yes, we can express "s" in terms of the circumference which is dependent upon "R". I believe it is fine to do this as you know that the particle has moved one-fourth across the circle.

    1. e^t is the component of movement or acceleration that acts tangential to the object. e^n is the component that acts normal, or perpendicular to the motion of an object.

  3. e^t is a unit vector (vector of length "1") that is tangent to the path. Similarly, e^n is a unit vector that is normal (perpendicular) to the path. You are more familiar with the orthogonal unit vectors i_hat and j_hat that are used in the Cartesian description for kinematics. e^t and e^n are the unit vectors that are used in the path description for kinematics.

          1. Oh, my bad. I see. So technically speaking, we after we calculate v_dot, we can use that s value from arc length to get a scalar value for v_dot at B, which we can then use to get the acceleration at B?

                1. I am not sure of your question. The speed of the automobile at ANY location on the path for s > 0 (which includes point B) depends of the speed of the automobile at point A through the equation: v = v_A*cos(bs). You are given the speed of the automobile at A; you will use that to find the speed and the rate of change of speed of the automobile when it is at B.

                  1. Sorry, you answered it! I was trying to figure out the speed at v_B, which I used at s with the equation given.

                    One more question, if you don't mind. When taking the magnitude of acceleration, we keep it in terms of e_t and e_n as our x^ and y^, right? So the magnitude would be the sqrt(e_t^2 + e_n^2)?

  4. In the acceleration equation (v_dot*et + (v^2/p)*en) I am confused about taking the time derivative of the velocity function given. There is no explicit t term in this equation so does the whole equation act as a constant and the derivative is equivalent to 0 or is the a term (like s) that can be evaluated in terms of time? If it is the later case would you do that?

    1. As mentioned in the above discussion, since v is an explicit function of "s" instead of being an explicit function of "t", then you need to use the CHAIN RULE of differentiation to find dv/dt:
      v_dot = dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds).

    2. Since the velocity equation V is in terms of S you have to do a similar process to the first homework where you are ultimately trying to get the derivative with respect to time. Since S is in terms of time you can use chain rule to take dv/ds * ds/dt and now the velocity equation derivative is with respect to t.

    1. I would try checking if your units are correct for the v_dot_B calculation and that you got a negative number (because the driver hits the breaks), if you used the chain rule incorrectly you may have ended up with something other than m/s^2 or get a positive result. You could also convert v_B to mph to check that it is a fairly realistic speed for a car.

Leave a Reply