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 e_t  and a = v_dot e_t + (v²/ρ) e_n.

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??

Discussion and hints:

In this problem, we know both the magnitude and direction for each of the velocity and acceleration vectors for the center of mass G. We know that the unit tangent vector e_t is in the same direction as the velocity vector v_G. The unit normal vector e_n is perpendicular to e_t, and points “up and to the right” (do you know why?). Projecting the acceleration vector a_G onto the unit tangent and unit normal vectors gives us the information that we need to find the rate of change of speed and the radius of curvature of the path.

Carefully watch the animation below of the motion of the aircraft. When the acceleration vector shown in RED points “backward” from the direction of travel, the aircraft is slowing down; that is, the rate of change of speed is negative. Conversely, when the acceleration vector is “forward” of the direction of travel, the aircraft is increasing in speed with a positive rate of change of speed. If you follow this logic, you are on your way to understanding the usefulness of the results of the path description of kinematics!

Any questions??

Discussion and hints:

In this problem, one simply needs to differentiate both the x- and y-coordinates once with respect to time in order to find the corresponding Cartesian components of velocity. Taking another time derivative of each then produces the Cartesian components of acceleration.

The results of this analysis are shown in the animation below. From this animation, we see that the velocity of the P (shown in BLUE) is always tangent to the path of P. The acceleration (shown in RED) is not quite so simple: it has components both tangent and perpendicular to the path; however, the component perpendicular to the path always points inward on the path. Both of these observations are consistent with what we see in the next lecture when we develop the expressions for velocity and acceleration in terms of their PATH components.

Any questions??

Discussion and hints:

In this problem, the xy-Cartesian components are constrained through the equation of the path of P:  x² + y² = R².  The x-component of the position is an explicit function of time. Finding the time derivatives of x is done through simple time differentiation of x = b sinωt. The derivatives of y with respect to time is found through the time differentiation of the constraint equation: x² + y² = R². Please review the examples worked out in lecture to assist you in this differentiation.

The results of this analysis are shown in the animation below. From this animation, we see that the velocity of the P (shown in BLUE) is always tangent to the path of P. The acceleration (shown in RED) is not quite so simple: it has components both tangent and perpendicular to the path; however, the component perpendicular to the path always points inward on the path. Both of these observations are consistent with what we see in the next lecture when we develop the expressions for velocity and acceleration in terms of their PATH components.

Any questions??