The following are links to hints and suggestions on the conceptual problems that appear at the end of each chapter in the lecture book. Note that for virtually all of these questions we are not expecting you to just “know” the answer based on your intuition. In each case, instead, you need to perform some simple analysis first, and from this analysis, draw conclusions on the answers.
Our hope is that through such conceptual questions your intuition on dynamics will develop beyond just remembering methods and procedures. You will see similar conceptual questions on your quizzes and exams in this course.
Use these hints to guide you in your conclusions. Please post your comments and questions below regarding these conceptual questions.
I was wondering for C1.7 if the answer is decreasing in class we got increasing but when I am redoing it I am getting decreasing when I am projecting the vectors.
Also, for C1.10 there is a note saying the relative velocity equation doesn’t hold because one plane is rotating, but I thought those equations always held. So is ‘both (a) and (b)’ not the correct answer?
For C1.10:
You are correct in that v_A/B is always the velocity of A relative to B, and v_B/A is always the velocity of B relative to B.
However, v_A/B is equal to to velocity of A AS SEEN BY OBSERVER B only if B is not rotating, and v_B/A is equal to to velocity of B AS SEEN BY OBSERVER A only if A is not rotating. Since B is rotating as it moves on its path, then v_A/B is NOT equal to to velocity of A as seen by observer B.
Does that help?
For C.17:
If you take the dot product of acceleration a_P with the path unit vector e_t you will get the negative number of -(6)*(10)/sqrt(40). Since that dot product represents the rate of change of speed, the speed of P is DECREASING. Therefore your answer is consistent with that result.
If a no-slip disk is in contact with an accelerating floor or moving platform, does the acceleration of the contact point stop being purely toward the center and instead have both x and y components? In some problems, even when the disk’s center is accelerating, the contact point’s acceleration still points toward the center. In what situations does the no-slip contact point have two acceleration components instead of just a radial (toward-center) one?
The answer to your first question is YES.
Do not try to memorize results; instead, focus on understanding why these results occur. Remember to use the rigid body kinematics equations – they will give you the answer.
For C4.18, why is the linear momentum in the x-direction not conserved for A+B? There are no forces imparted in the x direction at the instant of the collision. Does it have something to do with the movement of B being constrained along a circular path, which changes the linear momentum to angular? This problem looks like C4.15, which says that angular momentum would be conserved for this scenario, disproving that explanation.
It appears that linear moment is in fact conserved for the system A + B in the x-direction, as the answer key says the the sum of the forces in the x-direction is equal to zero. The only force acting on the system is through the bar (Fob).
For the system made up of A+B I believe the momentum is conserved in the X direction because the sum of the forces in that system’s FBD is equal to zero (noted in the solutions). The bar exerts a vertical force on the system of A+B, making momentum not conserved in the Y direction (noted in solution as sum of forces not equal to zero).
For question C4.18, I’m confused why momentum is not conserved in the x-direction for the A+B system. According to the FBD given, the only force acting on the system is the applied by the rigid bar on the system, which is only in the negative y-direction. Since the sum of forces in the x-direction is zero, linear momentum should be conserved in the x-direction, yet the answer key says that linear momentum is not conserved in the x-direction.
You are correct in your arguments. We have fixed the error in that file.
Thanks for letting us know.
It’s an inelastic collision, so linear momentum is only conserved in the t direction, but not in the n direction(because it’s the line of action which imparts impact force) which does have an x componet.
*which is why it does not have an x componet because the forces cancel out.
For C5.10 does saying the wheel is slipping automatically mean you can assume friction is not static and has to be kinetic?
This is correct.
For C5.18, when it says relative speed of the block and sphere, is it referring to linear component of velocity only, or is it referring to which object crosses the distance d the fastest?
In this case of traveling over the same distance with each acceleration being constant and the same initial speed, those two things (velocity or crossing the finish line) are the same.
For Question C5.10, I’m unsure how to determine the direction of the friction force acting on the drum at C. How do we know which way the drum is moving?
I believe a good way to look at this would be to examine the setup as if there were no friction at the bottom of the drum. In that case, the drum would want to rotate counter-clockwise due to the cable. Then by reintroducing friction at the bottom of the drum, we know it would oppose the motion at the contact point between the drum and the ground, and since we know that the drum would tend to rotate counterclockwise causing the contact at the bottom to move to the right, friction would be acting to the left.
For C6.1 – When writing the spring force terms for particles A and B, how do you consistently determine the correct sign or direction of each force? Specifically for the middle spring between A and B? Does it always just oppose the direction of the pending motion?
What has helped me consistently when checking if my spring force directions is correct, is remembering that if the spring stretches, the force is going to be pointing towards the wall. If the spring compressed, it’ll be pointing toward the object. Spring force in between two moving objects point outward if compressed, and inward if stretched. But even if you get the directions of your spring forces wrong, as long as you do k(final x – initial x) the math should still work out
I also had just an overall question regarding the direction of friction at the contact point for a rolling disk. I was curious how we are supposed to determine the direction. Can it just be an assumed direction and then let the signs work themselves out in the math?
For C4.5 I’m confused about the geometry for delta as described in the hint. I understand that at position 1 we are a distance d away from b but I don’t understand how to find position 2. Would it just be -h and then we integrate it would become d-h?
For C5.17, can we say that since the mass moment of inertia is less for A than B, hence V_a2 will be less than V_b2 as they fall from the same height and have the same radius, so their omega values should be the same. yes?
Since the moment of inertia for A is more than it is for B, more energy goes into making the disk rotate than its translation, which is why V_a2 is less than V_b2.
For C6.7, an issue I ran into but later fixed was, when switching from x coordinate to z, use z = x-xst, as your just shifting your reference point to equilibrium. This shift gets rid of constant forces so the equation is much easier to use. Any term that is constant in x should disappear in the z form.
For C6.12, to find the phase relationship in an undamped system, compare the excitation frequency to the natural frequency, if the excitation frequency is greater than the natural frequency, the denominator when finding the amplitude becomes negative. This means that the response is 180 degrees out of phase.
For C4.16, would it be correct to assume that since 0<e<1, energy will be lost in the normal direction (so Vy2Vy2, which would result in a shallower exit angle? Or would a mathematical solution be required for this problem.