# What is the tangential component? Taylor p.347

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• SebastianRM
In summary, the centrifugal component of the acceleration due to gravity is given by the equation g = Ω^2 * R * sinθ * cosθ. The correct method to derive this component involves multiplying the magnitude of the centrifugal acceleration by the appropriate trigonometric function, which is not specified in the conversation.
SebastianRM
I do not understand why the tan component for a gravity affected by the centrifugal force:
g = Ω^2 * R * sinθ * cosθ

So I tried to draw this: using a "big" X-shaped axis where the / component goes along the main gravity direction while \ points normal to / this direction. Then the centrifugal force points out of the center of the X into the x direction, with and angle θ between the y-axis and the / direction. Thus the angle between the centrifugal force and \, should be 90 - θ, and so its projection along that direction should be cos(90-θ).
So i did not arrive to the conclusion of the book, where am I going wrong?

SebastianRM said:
I do not understand why the tan component for a gravity affected by the centrifugal force:
g = Ω^2 * R * sinθ * cosθ

So I tried to draw this: using a "big" X-shaped axis where the / component goes along the main gravity direction while \ points normal to / this direction. Then the centrifugal force points out of the center of the X into the x direction, with and angle θ between the y-axis and the / direction. Thus the angle between the centrifugal force and \, should be 90 - θ, and so its projection along that direction should be cos(90-θ).
So i did not arrive to the conclusion of the book, where am I going wrong?

I cannot tell you where you went wrong because your description of what you did is not clear to me. Here is my suggestion of how to derive what you need.

The centrifugal part of the acceleration according to equation 9.44 is ##~~\vec a_{cf}=\Omega^2~R~\sin\theta~\hat \rho##. Now look at figure 9.10. It shows the centrifugal acceleration horizontally out. The component you are seeking is tangent to the surface of the circle representing the Earth at that point. To get that component you need to multiply the magnitude ##a_{cf}## by the appropriate trigonometric function. What might that be?

SebastianRM

## What is the tangential component?

The tangential component, also known as the tangential vector, is a vector that is perpendicular to the normal vector at a given point on a curve or surface. It is used to describe the direction and magnitude of a change in a particular direction.

## How is the tangential component calculated?

The tangential component can be calculated using the dot product of the velocity vector and the unit tangent vector. This can be represented mathematically as T = v⋅T, where T is the tangential component, v is the velocity vector, and T is the unit tangent vector.

## What is the significance of the tangential component in physics?

In physics, the tangential component is important in understanding the motion of objects in circular or curved paths. It helps determine the direction and magnitude of the object's acceleration, as well as its angular velocity.

## Can the tangential component be negative?

Yes, the tangential component can be negative. This indicates that the object is moving in the opposite direction of the tangent vector, or that its velocity is decreasing in magnitude.

## How does the tangential component relate to the normal component?

The tangential component and the normal component are perpendicular to each other, forming a right angle. Together, they make up the total velocity vector of an object moving along a curved path.

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