Question

A uniform rod of mass M = 326 kg is in the shape of a semicircle of radius R = 6 m. Calculate the magnitude of the force on a point of mass m = 74 kg placed at the center of the semicircle, as shown in the figure.

Answer 1

The angular mass density is M/pi, the mass between theta and theta + dtheta is thus M/pi dtheta. If the two ends of the semiciscle are at -pi/2 and pi/2, then the foce will be in the theta = 0 direction. The component in that direction from a mass element at theta is obtained by multiplying the force from that direction by cos(theta).

The magnitude of the force is thus given by:

M/pi m G/R^2 Integral from -pi/2 to pi/2 of cos(theta) dtheta =

2/pi M m G/R^2

Answer 2

Thank you very much.

Answer 3

I appreciate.