Homework H1.D – Sp 26

9 Comments
Problem statement
Solution video

DISCUSSION THREAD

Discussion and hints:

Here, the distance traveled along the circular path, s, is known as an explicit function of time. The speed of P at any instant is simply v = ds/dt, and the rate of change of speed is: v_dot = d2s/dt2. Since the path is circular, the radius of curvature is given by: ρ = R. With these three quantities, you have everything that you need to write down the velocity v = v et  and a = v_dot et + (v2/ρ) en.

Carefully watch the animation below of the motion of P. When the acceleration vector shown in RED points “backward” from the direction of travel, particle P is slowing down; that is, the rate of change of speed is negative. Conversely, when the acceleration vector “forward” of the direction of travel, P is increasing in speed with a positive rate of change of speed. You will see that the period of time during which the speed of P is increasing is rather short, occurring during a time immediately after the direction of P changes.

Any questions??

9 thoughts on “Homework H1.D – Sp 26”

  1. When we sketch the acceleration vector, are we supposed to sketch 2 different vectors? One in the et direction and one in the en direction? or are we supposed to sketch them as one combined vector?

  3. When it asks to show the position of P, do we just need to draw it along the circular path or does it want us to calculate the distance from the left of the path in radians or something else?

    1. We are asking for just a sketch of where P is at that instant. Calculate the value of “s” and determine where along the path that location is. Make of sketch of the quarter circle, showing roughly where P is. At that location, show e_t and e_n, along with v and a., again as sketches.

  4. Can we leave our solutions in terms of e^c where c is a constant or should we use a calculator to solve for exact values? Also if we do solve for exact values how many significant figures should we use?

Leave a Reply