The 60kg collar slides on the curved rod in the vertical plane with negligable friction under the action of constant force F in the cord guided by the small pulley at D. If the collar is released from rest at A, determine the force F which will result in the collar striking the stop at B with a velocity of 4m/s. Point A is located 0.6 m horizontally away from point B and 0.2 m below the pulleys at D. Point B is located 0.2 m above and to the left the pulleys at D.

I know that I have to put this into cylindrical coordinates but I have no idea where to start or how to do that. Does anyone have any idea?

To solve this problem using cylindrical coordinates, we can convert the given information into the cylindrical coordinate system, which consists of the radial distance (r), the azimuthal angle (θ), and the height (z).

Given:
- The collar has a mass of 60 kg.
- The collar slides on the curved rod in the vertical plane with negligible friction.
- The collar is released from rest at point A.
- Point A is 0.6 m horizontally away from point B and 0.2 m below the pulleys at D.
- Point B is located 0.2 m above and to the left of the pulleys at D.

To start, let's consider the vertical motion of the collar. When released from rest at point A, the collar will fall vertically downward due to the force of gravity (mg), where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).

Next, we need to determine the net force acting on the collar in the vertical direction that will result in the collar striking the stop at point B with a velocity of 4 m/s. We can break down this net force into two components:

1. Force due to gravity (mg):
Since the collar is sliding on the curved rod, the force of gravity acting on the collar (mg) will have two components: one along the rod and one perpendicular to the rod.

The component of mg along the rod (mgθ) will cause the collar to slide downwards, and the component perpendicular to the rod (mgn) will keep the collar in contact with the rod. However, since the friction is negligible, mgn can be ignored for now.

2. Force applied by the cord (F):
The force F applied by the cord will also have two components: one along the rod and one perpendicular to the rod.

The component of F perpendicular to the rod will counteract the component of mg perpendicular to the rod, while the component of F along the rod will contribute to the net force acting on the collar. We need to find this component of F which will result in the collar striking the stop at point B with a velocity of 4 m/s.

In cylindrical coordinates, the vertical equation of motion becomes:

m * (acceleration along z-axis) = (net force along z-axis)

m * (d^2z/dt^2) = Fz - mgθ

Since the collar is sliding vertically with negligible friction, we can simplify this equation to:

m * (d^2z/dt^2) = Fz - mg

Now, we need to determine the values of Fz and θ in terms of the cylindrical coordinates. The curved rod is in the vertical plane, so the angle θ is the azimuthal angle, which remains constant throughout the motion.

Next, consider the position of point A relative to the pulleys at D. Point A is located 0.6 m horizontally away from point B and 0.2 m below the pulleys at D. This means that the collar, at point A, is at a distance of 0.6 m from the pulleys and 0.2 m below them in the vertical direction.

Finally, to find the force F required to result in the collar striking the stop at point B with a velocity of 4 m/s, we need to solve the equation of motion using the given information.

To solve this problem using cylindrical coordinates, you will need to first establish the equations of motion for the collar and then solve for the force F.

Step 1: Establishing the equations of motion

Let's define a coordinate system with the z-axis pointing vertically upward and the ρ-axis pointing radially outward from the center of the curved rod.

The motion of the collar can be described by the forces acting on it. There are two forces at play:

1. The weight of the collar, which acts vertically downward with a magnitude of 60 kg multiplied by the acceleration due to gravity (9.8 m/s^2).
2. The tension T in the cord, which acts tangentially along the curved rod.

The radial component of the tension, Tϕ, does not contribute to the motion since there is no radial acceleration.

The axial component of the tension, Tz, provides the necessary force to accelerate the collar.

Considering the geometry of the system, we need to calculate the angle θ that the cord makes with the horizontal (horizontal plane):

θ = arctan((0.2-0.6)/0.2) = arctan(-2) ≈ -63.43°

Step 2: Applying Newton's second law

In cylindrical coordinates, we can express Newton's second law as:

maρ = Tϕ (1)
maz = Tz - mg (2)

Note that the mass of the collar (m) cancels out in both equations because it appears on both sides of the equation.

Step 3: Decomposing the equations

The radial component, Tϕ, can be expressed as:

Tϕ = T sin(θ)

The axial component, Tz, can be expressed as:

Tz = T cos(θ)

Plugging these expressions into eq. (2), we get:

T cos(θ) - mg = maz

Step 4: Solving for acceleration (az)

Since the stop at B is located 0.2 m above the pulleys, the displacement in the z-direction, Δz, is given by:

Δz = -0.2 m

The final velocity of the collar, vf, in the z-direction is given as 4 m/s.

Using the kinematic equation:

vf^2 = vi^2 + 2aΔz

where vi is the initial velocity, which is 0 m/s since the collar is released from rest, we can solve for az:

az = (vf^2 - vi^2) / (2Δz) = (4^2 - 0^2) / (2*(-0.2)) = 40 m/s^2

Step 5: Solving for force F

Plugging in the values into eq. (1):

T sin(θ) = mρaϕ

Since the collar is released from rest at A, ρ = 0.6 m, and aϕ = 0 (no angular acceleration). Therefore, the right-hand side of the equation is 0.

We can now solve for T:

T sin(θ) = 0

Since sin(-63.43°) = -0.891, the equation can be written as:

T(-0.891) = 0

This implies T = 0, which means there is no radial force.

Plugging the values into eq. (2):

T cos(θ) - mg = maz

T cos(-63.43°) - (60 kg)(9.8 m/s^2) = (60 kg)(40 m/s^2)

Simplifying, we get:

T(-0.45399) - 588 = 2400

T(-0.45399) = 2400 + 588

T = (2400 + 588) / -0.45399 ≈ -6479.04 N

Since the force F is the same as the tension T, the force F that will result in the collar striking the stop at B with a velocity of 4 m/s is approximately -6479.04 N.