How to calculate the centrifugal impulse in the y direction?

Question

How to calculate the component of the centrifugal impulse in the y direction created through travelling along a semicircle from the x axis to the negative x axis, 1st and second quadrants? I am trying to compute the total impulse taking the changing direction of the acceleration into account as a mass m travels along a semicircle at constant velocity v. Need to find the total impulse in the y direction if the semicircle is in the first two quadrants about the origin. Need equation in terms of velocity, radius, mass, time to complete semicircle path.

To calculate the component of the centrifugal impulse in the y-direction, we'll break it down into smaller parts. Assuming a constant velocity v and mass m, we'll use the following steps:

1. Find the centripetal acceleration (ac) in terms of v and radius (r): ac = v^2/r
2. Since the mass is moving in a semicircle, the acceleration vector will change direction. We'll resolve the acceleration into its x and y components using trigonometry.
3. Calculate the impulse (J) in the y-direction by integrating the y-component of the acceleration over time.

Here's the equation:

J_y = ∫[0,π] (m * v^2/r * sin(θ)) dθ

where θ is the angle from the x-axis to the position of the mass, ranging from 0 to π (covering the first two quadrants).

Simplifying and evaluating the integral, we get:

J_y = (2 * m * v^2)/r

This is the total impulse in the y-direction.

Is this correct?

Answer 1

You have calculated the integral right but the selected integral is totally wrong. You should calculate the time integral of the vertical force component, the variable shouldn't be the angle.

Of course you can make substitution so that the new variable is the position angle (dt=dθ/angular velocity), but there's no need to do it, because the result of the integration can be seen easily otherwise.

The mass particle moves in the beginning upwards. In the end of the calculation period it moves as fast downwards. The impulse = The change of the momentum = 2mv and that's downwards.

The integration of the vertical force component, of course, gives the same (checked; only divide your original integration result with the angular velocity, let it and the radius generate v to the denominator and eliminate the exponent of v).