# Find the center of mass of a uniform semicircular plate of radius R

• annamal
In summary: M}{\pi*R^2}dm = M*\frac{d\theta}{\pi}r = R(cos(\theta)\vec i + sin(\theta)\vec j)rcm = \frac{2*R}{\pi}\vec j$$In summary, the equation for finding the center of mass for a circular arc is not correct. annamal Homework Statement Find the center of mass of a uniform this semicircular plate of radius R. Let the origin be at the center of the semicircle, the plate arc from the +x axis to the -x axis, and the z axis be perpendicular to the plate. Relevant Equations ##rcm = \frac 1 M*\int(r*dm)##$$rcm = \frac{1}{M}\int_0^\pi(rdm)dm = \sigma{dA}dA = (\pi*R^2)*\frac{d\theta}{2\pi}\sigma = \frac{M}{\frac{\pi*R^2}{2}}dm = M*\frac{d\theta}{\pi}r = R(cos(\theta)\vec i + sin(\theta)\vec j)rcm = \int_0^\pi{\frac{R}{\pi}(cos(\theta)\vec i + sin(\theta)\vec j)} = \frac{2*R}{\pi}\vec j$$Where did I go wrong? Last edited: annamal said: Relevant Equations:: ##rcm = \frac 1 M*\int(r*dm)## Where did I go wrong? Right there. It is not ##\int r.dm##. vanhees71 haruspex said: Read the text there carefully. The r in that formula is a position vector, (x, y, z). So the equation becomes ##(x_{cm}, y_{cm}, z_{cm})=\frac 1M\int(x,y,z).dm##. That works because all the x are parallel. The r in your usage are not all parallel. Yes, so ##\vec r = R(cos(\theta)\vec i + sin(\theta)\vec j)## I meant ##\vec r_{cm} = \frac 1 M*\int(\vec r*dm)## What's the problem? annamal said: Yes, so ##\vec r = R(cos(\theta)\vec i + sin(\theta)\vec j)## That's only true of a point on the circumference, so you have found the mass centre of a semicircular arc. Lnewqban haruspex said: That's only true of a point on the circumference, so you have found the mass centre of a semicircular arc. Ah, I see. This doesn't work either$$\sigma = \frac{2M}{\pi*R^2}dM = \sigma dAdA = \pi ydyy_{cm} = \frac{1}{M}\int ydM = \frac{\sigma*\pi}{M}\int{y^2}{dy} = 2MR/3

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## 1. How do you find the center of mass of a semicircular plate?

To find the center of mass of a semicircular plate, you need to use the formula: x = 4R/(3π). This formula takes into account the radius of the plate (R) and the constant pi (π) to determine the x-coordinate of the center of mass.

## 2. Can the center of mass be located outside the semicircular plate?

No, the center of mass of a semicircular plate will always be located within the boundaries of the plate. This is because the center of mass is the point where the weight of the plate is evenly distributed, and in a semicircular plate, the weight is concentrated towards the center.

## 3. How does the center of mass change if the radius of the plate is increased?

If the radius of the semicircular plate is increased, the center of mass will also shift towards the outer edge of the plate. This is because the weight of the plate is distributed over a larger area, causing the center of mass to move further away from the center.

## 4. What is the significance of finding the center of mass of a semicircular plate?

The center of mass is an important concept in physics as it helps determine the balance and stability of an object. In the case of a semicircular plate, knowing the center of mass can help in designing structures or objects that need to be balanced, such as a seesaw or a bridge.

## 5. Can the center of mass of a semicircular plate be located using calculus?

Yes, the formula for finding the center of mass of a semicircular plate (x = 4R/(3π)) is derived using calculus. By setting up an integral to calculate the moment of the plate about the x-axis and equating it to the total weight of the plate, the x-coordinate of the center of mass can be solved for using integration.

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