One end of a cord is fixed and a small 0.420-kg object is attached to the other

Question

One end of a cord is fixed and a small 0.420-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 ? = 21.0

Explanation / Answer

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. | |U | | O==> 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 ÷ cos 20° = 5.21447 N This is the tension if the object was not moving. Centripetal force = m * v^2/r = 0.5 * 8^2 /2 = 16 N Total tension in cord = 4.9 + 16 = 20.9 N The 20.9 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 = 0.5 * 9.8 * cos 20° The net force that is directed toward the center = 20.9 – 0.5 * 9.8 * cos 20° Radial acceleration = net radial force ÷ mass Radial acceleration = (20.9 – 0.5 * 9.8 * cos 20°) ÷ 0.5 = 32.6 m/s^2 The direction of the tangential acceleration is perpendicular to the tension force in the cord, so the tension has no component causing tangential acceleration. The weight has a component that is tangent to the circle. The tangential force = m* g * sin ? = 0.5 * 9.8 * sin 20° Tangential acceleration = tangential force ÷ mass = 9.8 * sin 20° = 3.35 m/s^2 Total acceleration = (32.6^2 + 3.35^2)^0.5 = 32.77 m/s^2 d) Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? NO The weight and centripetal force are not dependent on the direction the object is moving. So the radial and tangential acceleration are not dependent on the direction the object is moving.

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