Consider a stainless steel annular disk with an outer radius 68mm and inner radi
ID: 2193016 • Letter: C
Question
Consider a stainless steel annular disk with an outer radius 68mm and inner radius 7.4 mm. The mass of the disk is 1349 grams.The stainless steel annular disk is allowed to rotate on a frictionless table with the rotation axis at its center. The disk has a small cylinder rigidly mounted at the top concentrically. The cylinder's radius is 12.5mm, and the mass of the cylinder is negligible. A string is wrapped around the cylinder, and a hanging mass of 19.3 g is tied at the other end of the string. When the mass falls under gravity, it causes the stainless steel annular disk to rotate. Ignoring the string's mass, and assuming that the string's motion is frictionless.
How much distance has the hanging mass been falling by this time? (Considering time is 4.4 seconds after the hanging mass is released from rest)
Explanation / Answer
Tension in the string due to hanging mass = 0.0193 * 9.8 = 0.189 N so, torque on the system = Tension * radius of cylinder = 0.189 * 0.0125 = 2.96 * 10^(-5) N m mass density (suface ) of the disc = mass / (pi*(Router^2 - Rineer^2 ) = 9.4 *10^(-5) kg/mm2 so mass of disc with 68mm as radius = mass density * pi*68^2 = 1.365 kg inertia of this disc = mass*r^2 / 2 = 1.365 * 0.068^2 /2 = 3.156 * 10^-3 kg m2 mass of disc with 7.4 mm as radius = mass density * pi*7.40^2 = 0.0162 kg inertia of this disc = mass*r^2 / 2 = 0.0162 * 0.0074^2 /2 = 4.435 * 10^-7 kg m2 so inertia of the current disc = inertia of disc with 68mm as radius - inertia of disc with 7.4 mm = 3.156 * 10^-3 kg m2 now torque = 2.96 * 10^(-5) N m inertia = 3.156 * 10^-3 kg m2 so angular acceleration = torque / inertia = 9.4 * 10^-3 rad/sec2 so acceleration of the mass = angular acceleration * radius of cylinder = 9.4 * 10^-3 * 12.5 = 0.117 mm/sec2 time = 4.4 sec s= ut + (0.5) * a * t^2 so, distance travelled = 1.1326 mm