Question

4) (12pt) An unknown mass M is non-uniformly distributed along a semi-circular arc of radius R such that the linear density is given by (ф)-16 sin®. where Mo is a constant and the angle φ is measured counterclockwise with respect to the positive x-axis. (a) In terms of R and A. what is the mass of the ring? (b) Derive an expression for the gravitational force of the ring on the particle of mass m at the center of the ring, expressed in unit vector notation and depending on the given quantities, m, R Hint You might find this trig identity useful: sin'-t-cos(2a).

Answer 1

Given linear mass density mu (phi) = mu_0 sin phi (mu_0 is constant here) To calculate the mass of the ring we take an element of length dl and angular length d phi at an angle phi anti clock from + x direction So dl = R d theta Now mass of the element dm = mu dl dm = mu_0 sin phi d phi To calculate the total mass we integrate it from angle phi, phi = 0 degree to phi = pi integral^M_0 dm = integral^pi_0 mu_0 R sin phi d phi M = mu_0 R integral^pi_0 sin pi d phi \r\n M = mu_0 R [- cos phi]^pi_0 M = mu_0 R [- cos pi + cos degree] M = mu_0 R [+ 1 + 1] M = 2 mu_0 R This is the expression for mass of ring in terms of mu_0 & R (b) To find gravitation force on mass m due to ring,