Consider a stainless steel annular disk with an outer radius 62 mm and inner rad
ID: 1907601 • Letter: C
Question
Consider a stainless steel annular disk with an outer radius 62 mm and inner radius 7 mm. The mass of the disk is 1366 grams. Keep your answers to at least 4 significant digits. (a) What is the moment of inertia of the stainless steel annular disk (answer in kg x meters squared)? (b) 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 12.6 mm, and the mass of the cylinder is negligible. A string is wrapped around the cylinder, and a hanging mass of 19.5 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, what is the angular acceleration of the stainless steel annular disk (answer in rad/seconds squared)? (c) What is the angular speed of the stainless steel annular disk 5.6 seconds after the hanging mass is released from rest (answer in rad/s)? (d) At what speed is the hanging mass falling at this time (answer in m/s)? (e) What is the rotational kinetic energy of the stainless steel annular disk at this time (answer in J)? Thanks in advance! I am having particular trouble with (b).Explanation / Answer
Mass (m) = 1.349 kg Inner radius (r1) = 0.0074 m Outer radius (r2) = 0.068 m Cylinder radius (r3) = 0.0125 m Disk mass moment of inertia (MMI) = ½* m * ( r1^2 + r2^2 ) = 0.0031558 kg - m^2 Applied force (hanging mass) = 0.0193 kgf * 9.80665 = 0.18927 Newtons (N) Applied torque = 0.18927 N * 0.0125 m = 0.002366 N - m Rotational acceleration (a) of disk = torque / MMI = 0.002366 / 0.0031558 = 0.7497 ( rad/sec ) / sec Radians traveled by disk and cylinder in 4.4 seconds : = ½ * a * t^2 = 7.257 radians = ( 7.257 / ( 2 * pi ) ) 1.155 revs = 2 * pi * r3 * 1.155 = 0.0907 m ( 90.7 mm )