A slotted disk is rotating with constant angular velocity ω−−, as shown in the figure. A ball of mass m slides out from the center of the disk. Kinetic friction force F=μk|N| acts on the mass, where N is the normal force between the ball and the channel. Using the body coordinate system i^, j^ fixed to the disk, derive the equation of motion for the mass using r(t) as the generalized coordinate.
Note: Gravity is not acting on this system.
Please enter symbolically the equation of motion in terms of r, r˙, m, ω, and μk.
In the co-rotating frame, define an angular variable alpha such that the channel is at alpha = 0. You can enforce that the ball is at alpha = 0 using a Langrange multiplier, so you treat the alpha coordinate as a dynamic variable for the ball.
The kinetic energy is:
T = 1/2 m r-dot^2 + 1/2 m r^2 (omega + alpha-dot)^2
The Lagrangian is:
L = T - V - lambda alpha
where lambda is the lagrange multiplier that will make sure that the ball stays at alpha = 0. V is the potential energy, which is zero in this problem but you keep it in the equations as an unspecified function.
The derivatives of V would give you the force from the potential, which is zero, but this then allows you to put in the friction force by hand (friction force is, of course, not conservative and cannot be specified by a potential).
The value of lambda gives you the normal force from which you can compute the friction force. You can then add this friction force in the Euler Lagrange equations by hand to get the correct equation of motion.
The answer is not clear! Can you explain ?
To derive the equation of motion for the mass sliding out from the center of the slotted disk, we need to consider the forces acting on it. In this case, the only force acting on the mass is the kinetic friction force F.
The first step is to calculate the normal force N acting between the ball and the channel. Since the ball is sliding along the channel, the normal force N will be the centripetal force required to keep the ball in circular motion. We can express this force as N = mω^2r, where ω is the angular velocity of the disk and r is the distance of the ball from the center of the disk (generalized coordinate).
Now, let's break down the kinetic friction force F into its components along the i^ and j^ directions fixed to the disk. Since the ball is sliding out from the center, only the j^ component of the force will contribute to the motion. Therefore, F = -μkNj^.
Next, we need to relate the acceleration of the mass to the given conditions. Since there is no gravitational force acting on the system, the acceleration will be directed along the j^ axis. Therefore, the acceleration can be expressed as a = r¨j^, where r¨ is the second derivative of r with respect to time.
Now, we can apply Newton's second law in terms of the j^ direction:
ma = ΣFj^
m(r¨)j^ = -μkNj^
Substituting N = mω^2r, we get:
m(r¨)j^ = -μkmω^2rj^
Now we can equate the components of the equation:
mr¨ = -μkmω^2r
Rearranging this equation, we have:
r¨ = -μkω^2r/m
Therefore, the equation of motion for the mass sliding out from the center of the slotted disk is:
r¨ = -μkω^2r/m
This equation describes the acceleration of the mass as a function of its position and the given parameters: r (generalized coordinate), ω (angular velocity of the disk), m (mass of the ball), and μk (coefficient of kinetic friction).