Homework 01 Posted by D. Dane Quinn on Wed 27 of Jan, 2016 13:58 EST posts: 472 ☆ ☆ ☆ ☆ ☆ Now batting... Homework 01. Lyric of the dayThere is moonlight and moss in the trees, Down the Seven Bridges road...
Homework 01 Posted by D. Dane Quinn on Wed 27 of Jan, 2016 13:58 EST posts: 472 ☆ ☆ ☆ ☆ ☆ Now batting... Homework 01. Lyric of the dayThere is moonlight and moss in the trees, Down the Seven Bridges road...
Posted by David Lloyd on Sat 30 of Jan, 2016 17:49 EST posts: 8 ☆ ☆ ☆ On the second problem of HW1, am I ok if I use the small angle assumption? This is in regard to the deflection of k2 which connects to the wall and the outer radius of the disk. If I assume a small angle, I can say that the axis of the spring stretch is in the i direction, whereas if I don’t assume a small angle, the axis of stretching rotates.
Posted by D. Dane Quinn on Sat 30 of Jan, 2016 18:05 EST posts: 472 ☆ ☆ ☆ ☆ ☆ Actually, you don’t need to use a small angle assumption. The cable to which the spring is attached is wrapped around the disk, so that the deflection of the spring is always in the i direction, regardless of the angle of rotation.
Posted by D. Dane Quinn on Sun 31 of Jan, 2016 20:09 EST posts: 472 ☆ ☆ ☆ ☆ ☆ Hi... not necessarily. You should just treat them as separate values.
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 19:58 EST posts: 472 ☆ ☆ ☆ ☆ ☆ No, r1 and r2 are just arbitrary values for the inner and outer radii...
Posted by AndrewTully on Wed 03 of Feb, 2016 22:35 EST posts: 17 ☆ ☆ ☆ Dr. Quinn, Did you post the homework solutions for hw 01? I did not see them.
Posted by AndrewTully on Wed 03 of Feb, 2016 22:56 EST posts: 17 ☆ ☆ ☆ Dr. Quinn, For problem 6, above the left spring constant symbol, k1, what is X denoting? Is it a symbol of the direction of force from spring/damper onto the mass? Is it the change in length from equilibrium of the spring/damper? Same question for problem 5, where a fixed position is noted as X. For problem 2, in the problem statement are you asking us to find the deflection of the spring from the complimentary solution set, or in other words when the differential equation describing the the deflection of the springs is equal to zero? How can we apply the next statement in the problem description that in the absence of gravity the springs are unstretched in their equilibrium equation?
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 08:10 EST posts: 472 ☆ ☆ ☆ ☆ ☆ In both problems 5 and 6, the coordinate x describes the displacement of the object. In problem 5 it is measured from the configuration of the system when the force F vanishes, and in problem 6 it is measuring from the equilibrium position. In problem 2, I am asking for the stretch in the springs when the system is in equilibrium. I am also describing the initial stretch in each spring in the absence of the gravitational forces, which is zero.
Posted by AndrewTully on Thu 04 of Feb, 2016 18:44 EST posts: 17 ☆ ☆ ☆ Professor Quinn, In your solutions you related the the displacement of the blocks as x1=-r*theta and x2=r*theta can I relate them opposite of that ? So, I would think x1=-r*theta and x2=r*theta Thanks
Posted by AndrewTully on Thu 04 of Feb, 2016 18:59 EST posts: 17 ☆ ☆ ☆ A similiar questions for the angular momentum equation you derived the sum of moments to be (T1 r − T2 r) kˆ = IG¨θ k and I would derive it to be (T2 r - T1 r). I assumed this because T2>T1 and they are the same distance from where we summed the moments, and it agrees with your positive sign convection of theta . So, in other words, the two external forces applied on the disk are -T1 and -T2 in the j direction. If T2>T1 then the disk will tend to rotate in the ccw direction around the k hat direction. So, when I apply the angular momentum equation it yields T2 r - T1 r = IG ‘’θ k.
Posted by AndrewTully on Thu 04 of Feb, 2016 19:18 EST posts: 17 ☆ ☆ ☆ Dr Quinn, Sorry I’ve go another question as well, How did you use the 3 equations from newtons second law to get the denominator : theta dot = (W1-W2)r/(Ig + (m1+m2)r^2)
Posted by NicholasSchifer on Thu 04 of Feb, 2016 19:23 EST posts: 3 ☆ On the problem 1a solution, I am able to follow everything up to “Combining with the coordinate constraints, we may solve for theta double dot.” Then there is some magic that happens to get the resulting solution. No matter how I try to combine the equations, I am unable to get the same result. What do I do with x1 double dot, and x2 double dot?
Posted by AndrewTully on Thu 04 of Feb, 2016 19:27 EST posts: 17 ☆ ☆ ☆ I have the same question Dr. Quinn.
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 20:01 EST posts: 472 ☆ ☆ ☆ ☆ ☆ It’s a matter of solving the set of equations. I would take the second and third equations listed and solve for T1 and T2, then substitute back into the first equation shown. Finally, you can eliminate x1_ddot and x2_ddot by using the coordinate relations, which should provide the answer.
Posted by NicholasSchifer on Thu 04 of Feb, 2016 20:14 EST posts: 3 ☆ Problem 1a discussion: I did exactly as you described, but I do not see how the x1_ddot and x2_ddot cancel out. My best guess was to take the second derivative of x1 and x2, which is 0. When I do that, I am still lost on where (m1 + m2)r^2 in the denominator comes from.
Posted by AndrewTully on Thu 04 of Feb, 2016 20:04 EST posts: 17 ☆ ☆ ☆ Professor can you elaborate further?
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 20:16 EST posts: 472 ☆ ☆ ☆ ☆ ☆ You have five equations. Three from momentum balance and constraint equations. Then, there are five unknowns (T1, T2, x1, x2, theta). Eliminate everything except theta and solve.
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 20:32 EST posts: 472 ☆ ☆ ☆ ☆ ☆ The second derivative of x1 is related to theta as x1_ddot = r theta_ddot
Posted by Adam Fox on Thu 04 of Feb, 2016 20:39 EST posts: 3 ☆ Dr. Quinn, Could you please explain how the forces acting on the block in Problem 6 are all negative? Thanks.
Posted by D. Dane Quinn on Thu 04 of Feb, 2016 21:52 EST posts: 472 ☆ ☆ ☆ ☆ ☆ If x is positive then the forces are in the -i direction.