Suggestions:
- Note that all points on the cable move with the same speed and move through the same distance in a given time. This speed is given by R*theta_dot. Therefore, the magnitude of the acceleration of A is given by R*theta_dot_dot. The sign that you associate with this acceleration will depend on the coordinate that you assign to A. For example, if you use "y" to describe the position of A with positive being downward, then: y_dot_dot = R*theta_dot_dot. If defined positively upward, then: y_dot_dot = - R*theta_dot_dot
- The stretch in the spring is measured by R*theta. Therefore, the force on the cable due to the spring is given by k*R*theta. This force should be drawn to the RIGHT on the cable in the FBD since a positive theta has the spring in tension (a pull on the cable).
2 comments:
does the no-slip factor tell us that there is no tension force? or is the mg force and a tension force at the same point
There will be a tension in the cable as the system moves. The tension is not equal to the weight nor is the tension the same at the spring as it is at block A.
In deriving the EOM for this problem, be sure the follow the Newton-Euler guidelines: draw FBD's of each component (drum and block A) individually.
The FBD of block A shows the relationship between the tension, weight and m*a term for A.
The FBD of the drum shows the relationship between the tension at the spring, the tension at A and the I*alpha term.
Let us know if this does not help.
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