Consider the system shown in Fig. 2. A wheel whose moment of inertia is 6.0 kg-m
ID: 1516398 • 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 friction 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 stiddness of the cord and the inertia of the axle. FIG. 2: Wheel rotating about a horizontal axis directed out of the page. t = .02 sec. t = 12 sec. t = 58 sec. t = 92 sec. t = 154 sec.Explanation / Answer
for the block , force equation is given as
mg - T = ma
T = mg - ma eq-1
Torque equation is given as
Tr - 0.8 = Ia/r
T(0.02) - 0.8 = 6a/(0.02)
T(0.02) = 300a + 0.8
T = 15000 a + 40 eq-2
using eq-1 and eq-2
15000 a + 40 = mg - ma
15000 a + 40 = 5 x 9.8 - 5 a
a = 0.0006 m/s2
Vo = initial velocity = 0 m/s
t = time taken
d = distance dropped = 1 m
using the equation
d = Vo t + (0.5) a t2
1 = 0 t + (0.5) (0.0006) t2
t = 58 s