Pendulum Drop, Nail, Circular Motion, Energy Name: an P. Status: ducator Grade: -12 Location: N/A Country: N/A Date: 11/16/2005 Question: I got an energy question for you... At what distance (x), below the suspension point, do you need to put a nail such that when you drop a pendulum of length (l), from the horizontal, it will swing around, hit the nail and then swing in a complete circle? Replies: Dan, Consider the circumstances. The problem is both energy and centripetal force. When the ball on the string reaches the top of the circle, it must have the proper net force if it is to be traveling in a circle: (mv^2)/r. When the ball reaches the top of the string, it will have both kinetic and potential energy. Zero velocity is too high. The first limit is force. In this problem, r is the radius of the final circle, the distance from the nail to the ball. Gravity can provide some of the force, but the string can provide only downward force. One necessity is that Gravity not provide too much force: mg < (mv^2)/r. The second limit is energy. Conservation of energy relates the initial height to the final height. For convenience, let us use the nail as zero height. Draw a picture of the apparatus at initial height (include the nail). Draw a picture when the ball reaches the lowest point, with the string just touching the nail. Draw a picture with the ball at the top of the circle, above the nail. | From the second picture, you can see that the initial height above the nail is (l-r). We know the initial speed is zero. Initial energy: E1=mg(l-r). In the final picture, height above the nail is (r). Final POTENTIAL energy is mgr. As for kinetic energy, we have only (1/2)(mv^2). Final energy: E2=(mv^2)/2 + mgr. Since energy is conserved in this problem, E1=E2: mg(l-r)= (mv^2)/2 + mgr We now have enough information to find r and v. First, use the energy relationship to solve for mv^2: mv^2=2mg(l-2r). Substitute this formula into the centripetal force relationship: mg < 2mg(l-2r)/r. Now, solve for r: 1 < 2(l-2r)/r r < 2l-4r 5r < 2l r < (2/5)l You can use the second picture (the ball at the bottom) to see just how the setup will look. If the ball makes contact with anything during the fall, or if the ball is too light or too big, energy will be lost during the fall. A lower nail may be needed to compensate for non-conservative forces. Dr. Ken Mellendorf Physics Instructor Illinois Central College Click here to return to the Physics Archives

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