6-1. Conservation law of mechanical energy

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1 6-1. Conservation law of mechanical energy 1. Purpose Investigate the mechanical energy conservation law and energy loss, by studying the kinetic and rotational energy of a marble wheel that is moving along a path composed of a slope and a circle. 2. Theory The kinetic energy of a marble wheel that is rolling down a slope is the sum of the translation energy and the rotation energy, where and are its mass and momentum of inertia, and and are the linear and angular velocity. If friction is negligible, the sum of potential energy and kinetic energy is conserved. This implies that: (1) is satisfied, where is the height of the wheel from an arbitrary standard point. When we consider a marble that is at rest at y=h rolling to y=0, from equation (1) we obtain. The moment of inertia of a marble with radius is, and since, expressing and ω as a function of v in equation (1) we get: (2) Let s consider a marble that starts rolling at the position with zero velocity and barely passes through the top position of the circular track as shown in Fig 2. The mechanical energy at position and the velocity at the lowest position satisfy the following relations: 1) Mechanical energy at the top position The mechanical energy at the top position is where and are the linear and the angular velocities at the top position,, (3) and they satisfy, where is the radius of the circular track.

2 If we consider the situation where the marble barely passes through the position, in that point the centripetal force is equal to the gravitational force: If we substitute equation (4), and in equation (3), we obtain (4). (5) Since the mechanical energy at the starting point mechanical energy at the position, we get has to be equal to the. (6) 2) Velocity at position. From the conservation of mechanical energy at position, where and are the linear and the angular velocities at position. Substituting and, we obtain (7) and if the marble barely passes through the track, also equation (6) has to be, (8) satisfied, so substituting equation (8) into equation (6):. (9) 3) Relation between the velocity at the position and the velocity at the position get: From the conservation of mechanical energy between positions and we (10)

3 where is the angular velocity at position, and is the height of point C with respect to the standard point( defined in B. Rearranging equation (10), we obtain 4) Parabolic motion. (11) The marble leaving the final point falls down following a parabolic arc. Let s define the origin where the vertical line crossing the final point meets the ground (refer to Fig 2), so that the line parallel with the ground is the -axis, and the perpendicular line is the -axis. Then the trajectory of the fall is, (12) where is the acceleration of gravity, is the initial point and is the initial angle. If we express the final position at the end of the fall as, from equation (12), we get: If we solve the above expression with respect to we get:. (13). (14) 3. Equipment

4 Figure 1 Figure 2

5 4. Method (1) Set up the equipment as in Fig 1. Make the final point horizontal and measure the distance. (2) Measure the initial height that makes a marble barely pass through the point. (3) Measure 5 times the position where the marble hits the ground. (4) Calculate by using and. (5) Compare it with the theoretical value obtained from equation (2) (6) If the theoretical and the experimental values are different, discuss it and calculate the loss of mechanical energy. (7) Set up the equipment such as in Fig 2. Make the final point C have the angle, repeat (2) and measure the new height of the final point. (8) Check whether satisfies equation (6) (9) Measure 5 times the position where the marble hits the ground. (10) Substitute the measured values of in equation (14), obtain the experimental value and substitute it in equation (11) to obtain the experimental value of (11) From equation (9) calculate and compare the result with its measured value. If they are different, discuss why. And as you consider, calculate the theoretical value of again and compare it with its experimental determination.

6 Data sheet (1) Measurement of the distance. (2) Measurements of the height and of the angle (3) Calculate the experimental values A. Measured velocity at position B. Theoretical velocity at position C. Error factor of velocity at position D. Energy loss E. Calculate the proportion between and and compare it with equation (6) F. Calculate (experimental) by applying equation (13) and (11). G. Calculate (theoretical) by applying equation (9). H. Determine the error factor of.

7 6-2. Collision and rotation energy (sweet spot) 1. Purpose By using a ruler and a glider, figure out how the final rotation energy of the ruler varies with the impact position with the glider in the case of a perfect elastic collision. 2. Theory When tennis and baseball players hit a ball, the rotation energy of the bat or of the tennis racket is transferred to the ball. If we simplify the situation, the motion of the bat or of the racket can be considered as a simple rotation. The fraction of the bat s or racket s rotation energy transferred to the ball depends on the distance between the axis of rotation and the impact point. In this situation, we call the position that maximizes this fraction sweet spot. We will indicate this point as SS1. Figure 1 When you hit a ball with a baseball bat, there exists an impulsive point that makes no impact on you, i.e. as you hit the ball the bat does not exert back any force on your hand. This point is generally different with SS1, and is called percussion point. Let s call it SS2. SS2 is also referred to as the zero-impulse sweet spot. The position of SS2 is given by the following equation:

8 (1) Here is the moment of inertia, is total mass and is the distance of the center of mass from the axis of rotation. For example, for a rod whose length is and the density is uniform SS2 is We can calculate SS1 and SS2 using a computer program. Alternatively, we determine them experimentally: for instance, hitting a ball with a baseball bat SS2 is the point that minimizes the impact on the hitter s wrist. Using a rod and a glider, (Fig 1) we can find SS1 by observing how far the glider travels up the slope after being hit, as we change the impact point between the rod and the glider: the position of the impact point that maximizes the distance travelled by the glider is SS1. Knowing such distance we can calculate the transferred energy: if the angle of the slope is and the distance travelled by the glider is, the transferred energy is just the kinetic energy of the glider immediately after the collision, which is also equal to its maximal potential energy as it climbs up the slope: (2) 3. Equipment

9 4. Methods (1) Set up the equipment such as Fig 1. (2) Turn on the air blower. Let the ruler fall and collide on the glider. Write down and (3) Increase in steps of 10cm, from 10cm to 90cm. (4) For each value of y repeat the experiment several times and calculate the average of. (5) Find SS1. (6) Calculate SS3 from equation (2). (7) If you have enough time, attach a mass of 100g to the ruler and repeat the experiment.

10 Kinetic energy General Physics Experiment, Department of Physics, Sogang University 5. Results -Ruler s mass : g -SS1 y= cm SS2 y= cm 6. Questions (1) Is it possible to make a super bat where SS1 and SS2 coincide in the same point? If it is possible, how can we do that for a bat with uniform density? (2) When we analyze this experiment, which assumptions are we making? What kinds of effects are caused by these assumptions?

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