One end of a cord is fixed and a small 0.560-kg object is attached to the other

Question

One end of a cord is fixed and a small 0.560-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.00 m, as shown in the figure below. When ? = 26.0

One end of a cord is fixed and a small 0.560-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.00 m, as shown in the figure below. When ? = 26.0 degree , the speed of the object is 8.60 m/s. At this instant, find the tension in the string. T = N At this instant, find the tangential and radial components of acceleration. ar = m/s2 inward At this instant, find the total acceleration. a total = m/s2 inward and below the cord at degree s Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? Explain your answer to part (d).

Explanation / Answer

The tension in the supports the weight of the object and provides the centripetal force that keeps the object moving in a circular path.

The vertical component of tension = T * cos 26?
The vertical component of the tension supports the weight of the object.
T * cos 26? = m * g = 0.560 * 9.81 = 5.49
T = 5.49 / cos 26? = 6.108184 N This is the tension if the object was not moving.

Centripetal force = m * v^2/r = 0.56 * 8.60^2 /1 = 41.41 N
Total tension in cord = 5.49 + 41.41 = 46.9 N
The 46.99 N tension force is directed toward the center of the circle. So, all of the tension is the force causing the radial acceleration.
The component of the weight that is directed toward the center = 0.56 * 9.81 * cos 26?

The net force that is directed toward the center = 46.9

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