Question

One end of a cord is fixed and a small 0.320-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 1.50 m, as shown in the figure below. When ? = 17.0

Answer 1

same solution but values change

Message: u mean the angle delta 20.0 degree not 5 20.08 so it should look like this. When angle delta = 20.0 degree ,the speed of the object is 8.00 m/s.

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U=delta

One end of a cord is fixed and a small 0.500-kg object is attached to theother end, where it swings in a sectionof a vertical circle of radius 2.00 m asshown in Figure P6.18. When u 5 20.08,the speed of the object is 8.00 m/s. At thisinstant, find (a) the tension in the string,(b) the tangential and radial componentsof acceleration, and (c) the total acceleration.(d) Is your answer changed if the object is swingingdown toward its lowest point instead of swinging up?(e) Explain your answer to part (d).

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 20?

The vertical component of the tension supports the weight of the object.

T * cos 20? = m * g = 0.500 * 9.8 = 4.9

T = 4.9