Consider the system shown in Fig. 2. A wheel whose moment of inertia is 6.0 kg-m
ID: 1511779 • Letter: C
Question
Consider the system shown in Fig. 2. A wheel whose moment of inertia is 6.0 kg-m^2 rotates about a horizontal axis. The function of the bearings produces a torque of 0.8 newton-m. The axle of the wheel has a diameter of 4 cm. A cord, wound several times around the axle, has at its free end an object of mass 5.0 kg. If this object is released from rest, how long will it take it to descend 1 meter? Neglect mass and -.Oddness of the cord and the inertia of the axle. FIG. 2: Wheel rotating about a horizontal axis directed out of the page.Explanation / Answer
given
I = 6 kg.m^2
frictional torque, fT = 0.8 N.m
hangung mass, m = 5 kg
diamater of axle, d = 4 cm
r = d/2 = 4/2 = 2 cm = 0.02 m
let a is the accaleration of the hanging block.
let T is the tension in the string.
net force acting on the block, Fnet = m*g - T
m*a = m*g - T
T = m*g - m*a
now use, net torque on the wheel, Tnet = I*alfa
(m*g - m*a)*r - fT = I*a/r
m*g*r - m*a*r - fT = I*a/r
m*g*r - fT = a*(I/r + m*r)
a = (m*g*r - fT)/(I/r + m*r)
= (5*9.8*0.02 - 0.8)/(6/0.02 + 5*0.02)
= 6*10^-4 m/s^2
let t is the time taken to move 1 m.
apply, d = 0.5*a*t^2
==> t = sqrt(2*d/a)
= sqrt(2*1/(6*10^-4))
= 57.7 s