### How Do You Calculate TheCenter Of Mass Of A Rod With A Ball Attached To The End?

__How Do You Calculate TheCenter Of Mass Of A Rod With A Ball Attached To The End?__

__Centre Of Mass__

Now let’s take an object, say a ruler, since this ruler obviously exists in space, it exists and has some mass. Our question here is – what is the centre of mass? The centre of mass is a point, and at that point, for dealing with this object as a whole or the mass of the object as a whole, we can pretend that the entire mass of the object exists at that point. To be clearer, let’s take an example, we’ll take one point on the ruler, somewhere around the centre as that’s where the centre of mass would typically be, and let’s just say that, this point is the centre of mass. Assume that the mass of the whole ruler is, like 10 kilograms, now if force is applied at the centre of mass, a force of 10N, then this ruler would accelerate the same exact way as would a point mass. This point mass could be a small of dot or a really small object weighing 10Kgs, here if we apply a force of 10N, this point would accelerate at the exact same speed as that ruler, that is at 1m/. It is important to note that this pointdoesn’t actually have to be in the object.

Now why is this information so useful to us? Sometimes we have some crazy, really bizarre object and we want to figure out exactly what these objects do and if we know its centre of mass first, we can know how that object will behave without having to worry about the shape of the object.

__Where Is The Centre Of Mass?__

Let’s look at an easy way to realise where the centre of mass in an object is. Now, if an object has a uniform distribution, meaning what it’s made out of and its density do not change throughout the object, then the centre of mass will be the object’s geometric centre. In this case, a ruler is almost a one-dimensional object, so the centre of mass will be somewhere right in the middle, equidistant from each side. Now if we take something two-dimensional like a triangle, it’s centre of mass will be the centre of two dimensions, again the geographical centre. Now if we take something like a square, with half of it filled with lead and the other half with Styrofoam. In this case the centre of mass isn’t going to be the geographic centre, since lead is denser than Styrofoam. So, the centre of mass will be somewhere near the lead as the whole object does not have uniform density.

Now, let’s take that ruler and try to apply force away from the centre of mass. What happens here is the whole ruler will not get pushed forward but instead the whole ruler will rotate around the centre of mass. Now for example, you throw a monkey wrench at something, the wrench will be spinning in the air before reaching your goal, mid-air there it’s spinning around its centre of mass. Meaning when an object is free, when its to fixed to any point it rotates around the centre of mass.

__Calculating The Centre Of Mass__

Now let’s use all that information as a basis and get to the question in hand and calculate the centre of mass of a rod with a ball attached to it. We’ll go through two different methods.

__Method 1:__

Here first, we have to find the centre of gravity (CG) relative to the ball. Let’s say rod length is L, rod weight is Lw and Bw is the ball weight. While the rod creates a moment of Lw * L/2, the ball crates no moment at its centroid. Lw + Bw is the total weight of the system. Hence, if we idealize the point mass as from the ball, the distance to the centre of gravity is,

(Lw * L/2) / (Lw + Bw)

Let’s find the centre of gravity relative to the opposite end of the rod. The rod creates a moment of Lw * L/2, and the ball creates a moment of Bw * L. Hence, the distance from the non-ball end to the centre of gravity is,

**(Bw * L+Lw * L/2) / (Lw + Bw)**

Now we can text the above given equation with an example. Let’s say we take a ball that weighs 16 lbs (think a 2×4 bowling ball on the end) of an 8 ft rod weighing 10lbs. Using the first equation, the distance from the system CG to the ball is,

(10 * 8/2) / (10 + 16) = 1.54 ft

By using the second equation, the distance from the CG to the non-ball end of the rod is,

(16*8 + 10 * 8/2) / (10 + 16) = 6.46 ft

And now the answer checks out because, 1.54 ft + 6.46 ft = 8 ft.

__Method 2:__

This is an easier and more clean-cut method, which is fun to do as well. Now from the ball-free end, grab the ball. Now allow the rod to swing like how a pendulum would and measure the period. Suggestion – measure the time for ten swings by finding the time for one swing and then dividing by ten. Note that the time from start-to-opposite end-back-to-start is the period of a pendulum.

One you determine the period, use the period formula,

**T = 2 pi sqrt length / g**

Note, T – period

pi – 3.1416

G – 9.81 m/s/s

Now, the point L (length) away from the axial point of the swing would amazingly be the location of the centre of gravity.

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