Discussion and hints:

This is a typical cable-pulley problem for us in this course. Here, we need to first write the length of the cable, L, in terms of the position variables of s_A and s_B, along with some constants. Since it is assumed that the cable does not break, go slack or stretch, the length L is constant. Take a time derivative of the expression above for L. This relates the speeds of blocks A and B.

Carefully study the animation of the motion of this cable-pulley system provided below. One of the questions asked of you is the speed of B when s_A = 0. What do you see in the animation below as s_A passes through zero? Does your equation above tell you the same thing?

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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.

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

DISCUSSION and HINTS

The problem is posed in terms of a mixture of path and polar descriptions. We are asked to find path variable descriptions for rate of change of speed and radius of curvature.

The two path unit vectors of e_r and e_theta are shown above in magenta in the above animated GIF. The velocity (in blue) is 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 are directly tied to the rate of change of speed of P, also as expected.

In terms of problem solving, the following steps are recommended:

• Show both sets of unit vectors in your sketch: e_t, e_n, e_r, e_theta.
• Using the sketches, determine numerical values for r_dot, theta_dot, r_dotdot and theta_dotdot, based on the known velocity v and acceleration a vectors.
• Take the dot product of a with e_t to find the rate of change of speed, v_dot.
• Use the magnitude of the acceleration vector to solve for the radius of curvature, rho.

Please note that "r" is radial distance between O and P, and it is NOT the radius of curvature for the path.

Discusssion

As you watch the sphere move through its path, focus on a few things:

• Direction of the velocity vector. The velocity of the ball is always tangent to the path, as expected. Note that the unit tangent vector e_t points in the same direction as the velocity.
• Direction of the acceleration vector. The acceleration of the ball always points "inward" on the path, again, as expected. When P is slowing down, you see that the acceleration vector points in the negative e_t direction. Similarly, when P is increasing in speed, the acceleration vector points in the positive e_t direction.
• Directions of the polar unit vectors. The e_r unit vector always points outward from  point O toward the sphere. The e_θ unit vector is perpendicular to e_r, and in the direction of increasing θ, as seen in the animation above.

Your task is to calculate R, R_dot and R_ddot using the given equation relating R and θ. This will require derivatives with respect to TIME; however, since R is given in terms of θ, you will need to use the chain rule of differentiation. For example: R_dot = dR/dt = (dR/dθ)(dθ/dt).

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Discussion

As you watch particle P move through its path, focus on a few things:

• Direction of the velocity vector. The velocity of the ball is always tangent to the path, as expected. Note that the unit tangent vector e_t points in the same direction as the velocity.
• Direction of the acceleration vector. The acceleration of the P always points "inward" on the path, again, as expected. Here you see that the acceleration vector points in the positive e_t direction; therefore, its speed must be increasing with time (as it is).
• Directions of the polar unit vectors. The e_r unit vector always points outward from  point O toward P. The e_θ unit vector is perpendicular to e_r, and in the direction of increasing θ, as seen in the animation above.

Your task is to calculate R and θ, and their time derivatives. Here, you are given R and θ explicitly as a function of time. Therefore, the chain rule of differentiation is not needed.

Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.

Ask and answer questions through the "Leave a Comment" link above. You can learn as much by answering the questions of others as you do from asking questions.

DISCUSSION:

As you know, the velocity of P is tangent to the path (where the path here is the coaster track), whereas the acceleration has components both tangent and normal to the path:

v = ve_t
a = v_dot e_t + (v^2/rho) e_n

You are provided equations for v, rho and theta in terms of the distance s traveled along the path. Therefore, you can simply substitute those expressions into the above equations for the velocity and acceleration vectors. The only remaining task is to find the rate of change of speed, v_dot. Since the speed v is given not as a function of time, but instead as a function of s, we need to use the chain rule of derivatives to find v_dot; that is:

v_dot = dv/dt = (dv/ds)*(ds/dt) = v*(dv/ds)

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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

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

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.

Ask and answer questions through the "Leave a Comment" link above. You can learn as much by answering the questions of others as you do from asking questions.

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.