A mass m attached to a string moves in a circle of radius R with velocity v. The string passes through a hole at the center. If the radius is reduced to R/4, what is the final kinetic energy?

Step 1: Identify the initial conditions and the principle governing the motion. Initially, the mass m moves in a circle of radius R_0 with velocity v_0. The initial kinetic energy is KE_0 = (1)/(2)mv_0^2. The string passes through a smooth hole, meaning the tension force acts radially and does no work on the mass. Therefore, the angular momentum of the mass about the center of the circle is conserved. Step 2: Apply the conservation of angular momentum. The angular momentum L of a particle of mass m moving with velocity v at a radius R is given by L = mvR. Initial angular momentum: L_0 = mv_0R_0. Final radius: R_f = (R_0)/(4). Let the final velocity be v_f. Final angular momentum: L_f = mv_fR_f. By conservation of angular momentum: L_0 = L_f mv_0R_0 = mv_fR_f Substitute R_f = (R_0)/(4): mv_0R_0 = mv_f((R_0)/(4)) Cancel m and R_0 from both sides: v_0 = (v_f)/(4) Solve for v_f: v_f = 4v_0 Step 3: Calculate the final kinetic energy. The final kinetic energy KE_f is given by: KE_f = (1)/(2)mv_f^2 Substitute the expression for v_f from Step 2: KE_f = (1)/(2)m(4v_0)^2 KE_f = (1)/(2)m(16v_0^2) KE_f = 16 ((1)/(2)mv_0^2) Since KE_0 = (1)/(2)mv_0^2, we can write: KE_f = 16 KE_0 KE_f = 8mv_0^2 Comparing this with the given options, option (3) is the correct answer. The final value of the kinetic energy is 8mv_0^2. Send me the next one 📸