torque=momentdisk*angular scceleration
moment disk=torque/(acceleration/radius)
torque=force*distance=1*9.8*radius
mmomentdisk=9.8(radius^2)/4.5m/s^2
A uniform disk of radius 0.45 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 1 kg object, as shown. When released from rest the object falls with a downward acceleration of 4.5 m/s^2. What is the moment of inertia of the disk? The acceleration of gravity is 9.8 m/s^2. Answer in units of kg m^2.
2 answers
Torque = I alpha
Tension in line = T
weight = m g = 1 * 9.8 = 9.8 N
so 9.8 - T = 1(4.5)
T = 5.3 N or kg m/s^2
torque = T * R = 5.3 R Nm or kg m^2/s^2
so
5.3 R kg m^2/s^2 = I alpha
but a = alpha R so alpha = a/R
so
5.3 R^2 kg m^3/s^2 = I kg m^2 * 4.5 m/s^2
I = 5.3 * .45^2 /4.5 = .212 kg m^2
Tension in line = T
weight = m g = 1 * 9.8 = 9.8 N
so 9.8 - T = 1(4.5)
T = 5.3 N or kg m/s^2
torque = T * R = 5.3 R Nm or kg m^2/s^2
so
5.3 R kg m^2/s^2 = I alpha
but a = alpha R so alpha = a/R
so
5.3 R^2 kg m^3/s^2 = I kg m^2 * 4.5 m/s^2
I = 5.3 * .45^2 /4.5 = .212 kg m^2
0 answers