a) A motor spins a disk (Idisk = 24.1 kg*m2) with a torque of 39 N*m, and a 18 N
ID: 1499784 • Letter: A
Question
a) A motor spins a disk (Idisk = 24.1 kg*m2) with a torque of 39 N*m, and a 18 N frictional force acts on the disk at a radius of 1.6 m. Attached to the disk at a radius of 44 m is a 40 kg mass. What is the moment of inertia of the system?
b) A motor spins a disk (Idisk = 25.9 kg*m2) with a torque of 39 N*m, and a 18 N frictional force acts on the disk at a radius of 1.3 m. Attached to the disk at a radius of 27 m is a 73 kg mass. What is the magnitude of the net torque in N*m of the system?
c) A motor spins a disk (Idisk = 25.9 kg*m2) with a torque of 38 N*m, and a 19 N frictional force acts on the disk at a radius of 1.3 m. Attached to the disk at a radius of 2.5 m is a 9 kg mass. What is the magnitude of the angular acceleration in rad/s2 of the system?
Explanation / Answer
(a)
Idisk = 24.1 Kg m^2
Imass = m*r^2
Imass = 40 * 44^2
Isystem = I disk + I mass
Isystem = 24.1 + 40 * 44^2
Isystem = 77464.1 Kg m^2
(b)
Torque due to Frictional Force, t = F*r
t = 18 * 1.3 Nm
t = 23.4 Nm
Net Torque, = 39 - 23.4 Nm
Net Torque, = 15.6 Nm
(c)
Idisk = 25.9 Kg m^2
Imass = m*r^2
Imass = 9 * 2.5^2
Isystem = I disk + I mass
Isystem = 25.9 + 9 * 2.5^2
Isystem = 82.15 Kg m^2
Torque due to Frictional Force, t = F*r
t = 19 * 1.3 Nm
t = 24.7 Nm
Net Torque, = Isystem * Angular Acceleration
38 - 24.7 = 82.15 *
= 0.162 rad/s^2
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