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DISCUSSION and HINTS

Although end B of the cable is moving with a constant speed, block A is not. Can you see this in the animation below?

Write down the length L of the cable in terms of the two position variables s_A and s_B. Since the cable does not stretch, set dL/dt = 0. (Take care in performing the derivative dL/dt.) This will give you an equation that relates the speeds of blocks A and B.

Does your answer show that v_A < v_B (a result that is indicated in the above animation)? In fact, your result should show that 2v_A < v_B: does it?

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DISCUSSION and HINTS
Since the cable does not stretch (R_dot = R_ddot = 0), an observer on aircraft A will see a circular path for B, although the true path of B is more complicated than that. Since θ_dot = constant, then θ_ddot = 0.

It is recommended that you write down the relative motion kinematics equations relating the motions of A and B:

v_B = v_A + v_B/A
a_B = a_A + a_B/A

and use a set of polar coordinates to describe v_B/A and a_B/A:

v_B/A = R_dot*e_R + R*theta_dot*e_θ = R*theta_dot*e_θ
a_B/A = (R_ddot – R*theta_dot^2)*e_R + (R*theta_ddot +2*R_dot*theta_dot)*e_θ = – R*theta_dot^2*e_R

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DISCUSSION
Shown in the animation below is the path taken by Particle P.

• As expected the unit vectors e_t and e_n are tangent and normal to the path of P, repectively.
• Also seen there is that the velocity vector is always aligned with the tangent unit vector, e_t, since v = v e_t.
• The acceleration vector, in general, has both tangential and normal components of acceleration. The normal component always points in the same direction as the unit normal vector, e_n. The acceleration vector either points “forward” of the motion (for which the rate of change of speed is positive; i.e., increasing is speed), or points “backward” of the motion (for which the rate of change of speed is negative; i.e., slowing down).

HINTS
Recall that since you do not know the y-component of motion of P explicitly in terms of time, you will need to use the chain rule to find dy/dt.

It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, e_t. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude!   e_t = v/v .

Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration.

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HINTS
It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, e_t. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude!   e_t = v/v .

Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration equation.

Do not confuse R with ρ – they are not the same thing.

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DISCUSSION and HINTS
The motion of P is shown in the animation below. As we saw in our last lecture, the velocity of a point is always tangent to the path of that point. The acceleration of a point will always have a normal component that points inward on the path. Can you make these general observations in the motion of P shown below? Can you tell from the animation when P is increasing in speed and when P is decreasing in speed? (Look at the direction of the tangential component of the acceleration relative to the direction of the velocity.)

Note that in this problem we are not given the radial variable r as an explicit function of time. Because of this, when we are finding dr/dt when need to use the chain rule of differentiation; that is, we should use: dr/dt = (dr/dθ)*(dθ/dt).

We encourage you to interact with your colleagues here on conversations about this homework problem.

With the polar description, we write the velocity and acceleration vectors for a point in terms of the polar unit vectors e_r and e_θ. These two unit vectors are shown in magenta in the animated GIF below.

• e_r points outward from the observer O toward point P
• e_θ is perpendicular to e_r and points in the positive θ direction.

The velocity (in blue) is seen to be tangent to the path, as expected, The  acceleration (in red) has both tangential and normal components: the normal component always points inward on the path, and the size and sign of the tangential component is directly tied to the rate of change of speed of P, also as expected.

In terms of problem solving, this problem is straight forward. Both r and θ are given as functions of time. The components of the velocity and acceleration vectors are found directly from the differentiation of r and theta with respect to time. The chain rule of differentiation is NOT needed here.

Some additional observations:

• Although r_ddot = constant, the r-component of acceleration is NOT constant.
• Although θ_ddot = 0, the θ-component of acceleration is NOT zero.

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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 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*e_t
a = v_dot*e_t + (v²/ρ)*e_n

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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*e_t
a = v_dot*e_t + (v²/ρ)*e_n

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 e_t and e_n 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 e_n direction at a  distance of ρ from P.

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Discussion

Consider the animation below of the motion for particle P:

• For all time, P moves with a positive x-component and a negative y-component of both velocity and acceleration. Why? Can you see this in the animation?
• As we will see in the next class period when we talk about the “path description” of kinematics, the velocity of a point is always tangent to the path of the point, and the acceleration always points “inward” on the path of the point. Do you see this in the animation?

The components of velocity and acceleration of P are found directly from the time derivatives of the position components:

v_P = x_dot i + y_dot j
a_P = x_ddot i + y_ddot j