Suggestions:
- FBDs: Draw individual FBD's of the drum and carriage.
- Impulse momentum equations: Use linear impulse momentum (LIM) equation (in x-direction, where x is in the direction of motion) for the carriage. Also, use both LIM equation in x-direction and the angular impulse momentum equation for the drum.
- Kinematics: Any needed?
- Solve
Let us know if you have any questions on this problem.
9 comments:
I am getting a fairly reasonable answer (i feel) for the v_2, but for omega_2 I am getting a larger number than the +3 rad/s given in the problem statement.
Intuitively, should you get a smaller number because resulting motion will be in the negative k direction?
Is anyone else having this problem?
Andy - did you use angular impulse momentum eqn? If you did, I believe you should get something like (Ho)1 = m(drum)*v1*d + I*w1, where d is the radius of the drum. (Ho)2 is basically the same, except v2 becomes d*w2, where w2 is your unknown. Integrate the moment caused by the tension force about O over the 10 seconds, then plug all the parts into the eqn. I'm not 100% sure I'm right on this, but I came up with a very reasonable answer for angular velocity (w2), ~-2.2rad/s I think? That makes sense b/c the drum starts spinning clockwise after the tension force is applied.
--to cdewes--
Be careful in writing down the angular momentum equation. It appears as if you are using Equation 6/15 (page 500) of the textbook: HA = IA*omega + m*vG*d. In this equation, d is the distance between the center of mass and point A. From your statement, it looks as if you are summing moments on the disk about point O (which is the center of mass of the disk). Therefore, d = 0. [You are using d to be the radius of the disk.]
My recommendation is to always use HA = IA*omega for a rigid body, and to use EITHER A = center of mass OR A = fixed point, as we discussed in lecture. With this, there is less confusion and is more closely connected with what we did earlier in the course with the Newton-Euler equations.
Let us know if this is not clear.
--to Andy--
It is always a good idea to check the reasonableness of your final result, as you are doing here.
In doing so, please keep in mind that the moment about O due to T is CW ("-") whereas the initial omega for the disk is CCW ("+"). Since the moment is negative, T will tend to decrease the value of omega:
IO*omega2 = IO*omega1 - integral(T*r*dt)
(potentially changing omega from "+" to "-"):
At time 2, the value (with sign) of omega should be less than the value of omega at time 1; however, the magnitude (absolute value) of omega2 could be larger than the magnitude (absolute value) of omega1.
Does this make sense?
I am a bit confused on setting up the angular and linear momentum equations for the drum and the linear momentum equation for the carriage. Would I include force 'T' as an impulsive force in LIM equation for the carriage? Similarly, would I include the velocity of the carriage in the LIM equation of the drum?
I was also wondering whether the LIM for the drum and carriage are conserved because of the constant force T?
The guiding principle is first draw your FBDs. Based on the FBDs, apply the LIM and AIM momentum equations. Include only forces that appear on each FBD in those equations.
In this problem it was suggested that you draw individual FBDs of the drum and carriage. In the drum FBD you will have the following forces: reaction forces at acting at O (call these Ox and Oy), the applied force T and weight. In the carriage FBD you will have the reactions forces at O (-Ox and -Oy), normal reactions at the wheels and weight.
Use the LIM equation (x) and AIM for the drum. The LIM equation will include Ox and T. The AIM about O will include only T.
Use the LIM equation (x) for the carriage. This equation will include -Ox.
Note that neither linear momentum nor angular momentum is conserved for either of these FBDs.
Does this help? If not, let us know.
Thank you professor, that help a lot.
Is there friction on the wheels on the carriage?
There's no friction explicitly stated, and with no coefficient of friction it'd be impossible to calculate.
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