A particle is released on the smooth inside wall of a cylindrical tank at A with a velocity v which makes an angle beta with the horizontal tangent. When the particle reaches a point B a distance h below A, determine the expression for the angle theta made by its velocity with the horizontal tangent at B. There is a constant radius r along this path.

I have no idea where to begin. Do you have to put this in cylindrical coordinates? If so how do you go about doing that?

To solve this problem, we can indeed use cylindrical coordinates. Let's first define our coordinate system for the cylindrical tank. The vertical direction is the z-axis, the radial direction is the r-axis, and the horizontal direction is the φ-axis (commonly used for cylindrical coordinates).

Now, we need to break down the velocity of the particle at point B into its radial and tangential components. This will help us determine the angle θ.

To begin, let's express the coordinates of point B in terms of cylindrical coordinates. The height of point B below point A is given as h, which is the distance along the z-axis. The particle is moving along the cylindrical wall with a constant radius r, which means the radial distance remains constant.

Using cylindrical coordinates, the position vector of point B can be written as:

r_B = (r, 0, -h)

Now, we can determine the velocity vector of the particle at point B. The velocity vector is the derivative of the position vector with respect to time. However, since the particle starts at rest at A, the magnitude of its velocity will be constant throughout its motion along the cylindrical wall.

The velocity vector has two components: the radial component (vr) and the tangential component (vθ). The tangential component is responsible for the change in the angle made by the velocity vector with the horizontal tangent.

Let's denote the angle made by the velocity vector with the horizontal tangent at B as θ. We can express vθ in terms of the given angle β and the constant velocity v.

vθ = v * sin(θ)

Next, we can relate the tangential and radial components of the velocity using the cylindrical coordinate system. The tangential component is related to the angular velocity (ω), which is the rate of change of the angle φ with respect to time. Since ω is constant in this problem, we can write:

vθ = r * ω

Now, we can equate the two expressions for vθ to find an equation relating ω and θ:

v * sin(θ) = r * ω

Simplifying this equation will give us the expression for the angle θ made by the velocity vector with the horizontal tangent at B.

To summarize, we need to express the position vector of point B in cylindrical coordinates, determine the tangential and radial components of the velocity vector, and equate the tangential component to find an equation relating ω and θ.

Yes, to solve this problem, it would be helpful to use cylindrical coordinates. Cylindrical coordinates consist of a radius (r), an azimuthal angle (θ), and a vertical component (z).

To begin, let's consider the motion of the particle in the plane. From the problem statement, we know that the initial velocity vector makes an angle β with the horizontal tangent. Let's consider this angle as measured counterclockwise from the positive x-axis.

We can then decompose the initial velocity vector into its horizontal and vertical components. The horizontal component will be equal to v * cos(β), and the vertical component will be equal to v * sin(β).

Next, let's analyze the velocity of the particle when it reaches point B. At this point, the particle will have moved a distance h below point A along the inside wall of the cylindrical tank.

To find the angle θ made by the velocity with the horizontal tangent at point B, we need to find the angle between the velocity vector at point B and the horizontal tangent.

To do this, we should consider the projection of the velocity vector onto the tangent line at point B. This projection will be in the direction of the particle's instantaneous velocity as it moves along the inside of the tank.

Let's call this projection vector V.

Now, the vertical component of the velocity vector at point B will be equal to the vertical component of the velocity vector at point A, as there is no force acting in the vertical direction. This means that the vertical component of V will also be equal to v * sin(β).

To find the horizontal component of V, we need to consider the change in angle between point A and B. As the particle moves a distance h downwards, it also travels along an arc of the cylinder with radius r. The angle of this arc, which we will call φ, can be found using trigonometry.

Using the relationship arc length = radius * angle, we have:
h = r * φ

Rearranging, we find:
φ = h / r

Therefore, the change in angle between the horizontal direction at point A and B is φ.

Now, we can use trigonometry again to find the horizontal component of V. We have:
cos(φ) = adjacent side / hypotenuse

In this case, the adjacent side is the horizontal component of V, and the hypotenuse is the magnitude of V.

We can rearrange this equation to solve for the horizontal component of V:
horizontal component of V = hypotenuse * cos(φ)

Finally, we can find the magnitude of V by using Pythagoras' theorem:
magnitude of V = √[(horizontal component of V)^2 + (vertical component of V)^2]

Now, we have the horizontal and vertical components of V, so we can find the angle θ using trigonometry. We have:
tan(θ) = vertical component of V / horizontal component of V

Rearranging:
θ = atan(vertical component of V / horizontal component of V)

So, the expression for the angle θ made by the velocity with the horizontal tangent at point B is θ = atan(v * sin(β) / (v * cos(β) * cos(h/r))).