How do we relate the speed of P in terms of W so that we can solve for w using the energy equation and then solve for vp
Because P is constrained along the ground, and B is going to be moving straight up, and the sling is of length L is both cases, You can find the IC of the sling itself connecting B and P using the geometry at the two positions. You can force the angular velocity of the sling to agree with B’s velocity, then solving for P. You need to do some trig in both locations but it works out.
How did you find the distance from B and P to the IC? Is the angle from IC to B to P equal to theta?
Use the known information on theta and L to first find the vertical distance between ground and point O. Then knowing that the IC is directly above P on OB, you can use trig to find those two distances.
Since the projectile has to slide along the ground before it takes off, how does that tether dragging behind the beam change how fast it’s actually moving?
The rope behaves as a rigid link. Consider using the usual rigid body kinematics equation for velocity.
If we have the velocity of B, is the following equation the correct way of solving for P? I have an I and J component for velocity for P, which doesn’t seem correct.
V_P = V_B + omega x r_P/B
Yes, your equation is fine, provided that you use omega_BP (an unknown). Your vector equation has two scalar components in terms of the unknowns of v_P (where the vector for v_P is along the ground) and omega_BP.
Am I able to do Newtons equations around a point that is not the center of mass? If I did that I could find the tension from P which might be helpful later
Yes, you could use Newton equations around any point given you have the relevant information, but it is likely not particularly helpful to find the tension of the sling for this problem. It’s also worth mentioning that the tension will change throughout the motion, in both direction and magnitude. The most straightforward, effective method seems to be using the conservation of mechanical energy, and the sling can be treated as a virtually rigid link for the kinematics of this problem. Hope this helps.
1
An additional note on this – I remember the professors explained that when using the mechanical energy equation, it is best to make the system as “large as possible,” including as many elements of the system as possible to remove internal interacting forces, such as the tension between B and P, in order to simplify calculations. I found that if you include all objects in the system, you can ensure there are no non-conservative forces at work.
1
While P is still sliding on the ground, should we treat the motion from the release position to theta=90 using conservation of energy for the whole system, or do we need to account for a ground reaction/constraint on P that changes the approach?
Yes, if you use the system of D, AB, P and the rope, mechanical energy is conserved. The ground reaction on P does no work on the system, so therefore it will not appear in the W/E equation.
I have the work-energy equation set up, but I’m stuck with two unknowns (vp and w). I know there is a kinematic relationship between the two but I am not sure what it is. Does anyone have a hint on how to connect them?
You can write two sets of kinematics equations relating v_b to v_o and v_b to v_p. The latter will give you two equations for the i and j components of v_b. Knowing that v_b is constrained to the j direction and v_p to the i direction, you will be able to write v_p in terms of w_ab.
2
I have the same question, would it be possible to assume that Vp=Vb?
That would not be a good idea. B and P are different distances from the IC of the cable. Therefore, they will have different speeds.
Should we track 4 different heights in this problem? A height for D at states 1 and 2 and then a height for G at state 1 and 2. Or can we just use the D states?
I shifted the COM of D to be at A for both states since it says that length AD is relatively small. I then set my datum at the origin O, which then simplified my calculation for gravitational potential energy a little bit.
For this problem should we consider block D to be a particle
0
yes
I’m very confused on how to relate Vp and Vb, as well as finding the rotational KE of the arm, considering it isn’t rotating about its center
For KE, you should be able to take it about O so you only have the Iw term and then relate Io to Ig using the parallel axis thm. For vp and vb, the rope is a rigid link so u can use rigid body kinematics (vp = vb + rxw)
Can we treat the sling like an inextensible cable? Also, can we treat the beam as rotating about one end?
Based on a previous instructor reply the sling is treated as an inextensible cable. The beam doesn’t rotate about its end but you can use the parallel axis thm to find its moment of inertia such that I = Icm + md^2, where d is the distance of the rotation axis from the center of mass.
Does D have a kinetic energy or just a potential one?
I believe D would have both KE and PE, however it should be a simpler equation since D can be treated as a particle.
When writing the potential energy terms, how do we determine which masses actually experience a vertical displacement between the two positions?
Both D and the center of the mass of the beam have a vertical displacement between the two states; D is treated as a particle and AD has negligible length, so D will have the same height as A. The CM of the beam has a visible vertical displacement because the beam rotates about a point 0.2L away from the CM.
You can see that masses D, B, and the center of gravity of the bar G all experience vertical displacements from either below or above ground to the same plane as G, which is on the datum line at zero. The values of the displacements can be found using the angle and respective lengths from point O. Hope this helps.
Will the solution video be posted here? Ive noticed they haven’t been posted for a while and I find them helpful for studying. Understandable if not they are likely not easy to make but just wanted to check in case
How do we relate the speed of P in terms of W so that we can solve for w using the energy equation and then solve for vp
Because P is constrained along the ground, and B is going to be moving straight up, and the sling is of length L is both cases, You can find the IC of the sling itself connecting B and P using the geometry at the two positions. You can force the angular velocity of the sling to agree with B’s velocity, then solving for P. You need to do some trig in both locations but it works out.
How did you find the distance from B and P to the IC? Is the angle from IC to B to P equal to theta?
Use the known information on theta and L to first find the vertical distance between ground and point O. Then knowing that the IC is directly above P on OB, you can use trig to find those two distances.
Since the projectile has to slide along the ground before it takes off, how does that tether dragging behind the beam change how fast it’s actually moving?
The rope behaves as a rigid link. Consider using the usual rigid body kinematics equation for velocity.
If we have the velocity of B, is the following equation the correct way of solving for P? I have an I and J component for velocity for P, which doesn’t seem correct.
V_P = V_B + omega x r_P/B
Yes, your equation is fine, provided that you use omega_BP (an unknown). Your vector equation has two scalar components in terms of the unknowns of v_P (where the vector for v_P is along the ground) and omega_BP.
Am I able to do Newtons equations around a point that is not the center of mass? If I did that I could find the tension from P which might be helpful later
Yes, you could use Newton equations around any point given you have the relevant information, but it is likely not particularly helpful to find the tension of the sling for this problem. It’s also worth mentioning that the tension will change throughout the motion, in both direction and magnitude. The most straightforward, effective method seems to be using the conservation of mechanical energy, and the sling can be treated as a virtually rigid link for the kinematics of this problem. Hope this helps.
An additional note on this – I remember the professors explained that when using the mechanical energy equation, it is best to make the system as “large as possible,” including as many elements of the system as possible to remove internal interacting forces, such as the tension between B and P, in order to simplify calculations. I found that if you include all objects in the system, you can ensure there are no non-conservative forces at work.
While P is still sliding on the ground, should we treat the motion from the release position to theta=90 using conservation of energy for the whole system, or do we need to account for a ground reaction/constraint on P that changes the approach?
Yes, if you use the system of D, AB, P and the rope, mechanical energy is conserved. The ground reaction on P does no work on the system, so therefore it will not appear in the W/E equation.
I have the work-energy equation set up, but I’m stuck with two unknowns (vp and w). I know there is a kinematic relationship between the two but I am not sure what it is. Does anyone have a hint on how to connect them?
You can write two sets of kinematics equations relating v_b to v_o and v_b to v_p. The latter will give you two equations for the i and j components of v_b. Knowing that v_b is constrained to the j direction and v_p to the i direction, you will be able to write v_p in terms of w_ab.
I have the same question, would it be possible to assume that Vp=Vb?
That would not be a good idea. B and P are different distances from the IC of the cable. Therefore, they will have different speeds.
Should we track 4 different heights in this problem? A height for D at states 1 and 2 and then a height for G at state 1 and 2. Or can we just use the D states?
I shifted the COM of D to be at A for both states since it says that length AD is relatively small. I then set my datum at the origin O, which then simplified my calculation for gravitational potential energy a little bit.
For this problem should we consider block D to be a particle
yes
I’m very confused on how to relate Vp and Vb, as well as finding the rotational KE of the arm, considering it isn’t rotating about its center
For KE, you should be able to take it about O so you only have the Iw term and then relate Io to Ig using the parallel axis thm. For vp and vb, the rope is a rigid link so u can use rigid body kinematics (vp = vb + rxw)
Can we treat the sling like an inextensible cable? Also, can we treat the beam as rotating about one end?
Based on a previous instructor reply the sling is treated as an inextensible cable. The beam doesn’t rotate about its end but you can use the parallel axis thm to find its moment of inertia such that I = Icm + md^2, where d is the distance of the rotation axis from the center of mass.
Does D have a kinetic energy or just a potential one?
I believe D would have both KE and PE, however it should be a simpler equation since D can be treated as a particle.
When writing the potential energy terms, how do we determine which masses actually experience a vertical displacement between the two positions?
Both D and the center of the mass of the beam have a vertical displacement between the two states; D is treated as a particle and AD has negligible length, so D will have the same height as A. The CM of the beam has a visible vertical displacement because the beam rotates about a point 0.2L away from the CM.
You can see that masses D, B, and the center of gravity of the bar G all experience vertical displacements from either below or above ground to the same plane as G, which is on the datum line at zero. The values of the displacements can be found using the angle and respective lengths from point O. Hope this helps.
Will the solution video be posted here? Ive noticed they haven’t been posted for a while and I find them helpful for studying. Understandable if not they are likely not easy to make but just wanted to check in case