![]() How to find center of mass by integration? Definition of Center of Mass in Physics.In this article, we are going to explain the center of mass, to derive its formula by integration and will discuss why is it very much important in Physics? If the body is symmetric and with uniform density, then that central point is the center of mass of the body. Now, there must a central point in a body. The mass of the entire body is the sum of the masses of those points. From this, and the definition of density, I can find the ratio of densities.Every body consists of large number of point masses. I will assume the handle is a cylinder with a diameter of 5.4 cm which gives it a volume of 980 cm 3. Big deal, right? Ok, let’s use this to estimate the ratio of densities for the handle and the head. Putting in my values for the distances, I get a handle to head mass ratio of 0.295. This gives the following ratio ( here is my best explanation of center of mass and torque): I will call the distance from the center of mass to the handle a distance s 1 and the distance to the head s 2. If the hammer was balanced at the center of mass, then the torque from the handle must be opposite the torque from the head. The best way to find the ratio of handle to head mass is to consider torque. With this I can solve for the ratio of masses for the head to handle. So now I have the location of the center of mass, the center of the handle and the center of the head. If I use estimated hammer dimensions from the movie, I will say the length of the head is 26.3 cm (with a height of 16 cm) with a handle length of 42.8 cm. Now that I have essentially just two point masses, I can write the total center of mass as: Instead of fitting a quadratic equation to the data, I will fit a linear function and the slope of this function will be the horizontal velocity. The big difference is that the acceleration of the center of mass should be zero. Now I can do the same thing while looking at the horizontal motion of the hammer. I could use the linear function that fits the data to find the location where the acceleration is around -9.8 m/s 2 - but just guessing, it looks like 0.75 of the way from the bottom to the top is pretty close. The horizontal axis is the fraction of the way from the bottom of the hammer to the top. Here is the vertical acceleration for the different part. ![]() When I plot that position, I can then find the acceleration. By calculating the distance from the top of the hammer to the bottom (this appears to change as the hammer rotates), I can then multiple by a factor (say 0.1) to get a vertical position 1/10 th the way from the bottom to the top. ![]() To do this, I will break the hammer into small sections. This means that I just need to plot different points along the hammer and find the vertical acceleration. What I want to look for is a location in between the top and the bottom of the hammer that has a vertical acceleration of -9.8 m/s 2. However, I will proceed anyway (as usual). Second, the hammer is angled towards the camera instead of the optimal viewing angle that showing the hammer from the side. First, there are only a few frames in which both the top and bottom of the hammer can be seen at the same time. Of course, there a couple of problems with this video clip.
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