Homework H1.A - Fa22

Problem statement
Solution video

DISCUSSION THREAD

Given: A particle P travels on a path described by the Cartesian coordinates of y = cx(b-x). The x-component of velocity for P is a constant.


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From the animation below, you see that the velocity of P is always tangent to the path, and that the acceleration always points inward of the path. In the next class when we discuss the path description of kinematics, we will discover that this is true in general.

Note below that in this case the acceleration of P is always vertical. Do you know why?

HINTS:
The path of P is not given in terms of x- and y-coordinates that are known explicit functions of time (instead, the y-coordinate depends explicitly on the x-coordinate). For velocity and acceleration, we need the derivatives of x and y with respect to time. How do we do this? Recall the chain rule of differentiation:

dy/dt = (dy/dx)*(dx/dt)

Use this in setting up the components of velocity and acceleration.

 

7 thoughts on “Homework H1.A - Fa22”

    1. You need to take the two time derivatives, one at a time.

      First, use the chain rule to find y_dot:
      y_dot = (dy/dx)*(dx/dt) = (d/dx)[c*x*(b-x)]*x_dot

      Next, take a derivative of y_dot with respect to time using the product rule:
      y_ddot = (d/dt)y_dot

      Don't forget that x_ddot = 0.

    1. c is 5/m is there because the m^-1 is there to cancel out the m unit of x it's multiplied with. That way when you multiply cx (now dimensionless) with b-x (meters), the units of y are meters

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