In the figure below, the hanging object has a mass of m 1 = 0.495 kg; the slidin
ID: 2262225 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding block has a mass of m2 = 0.795 kg; and the pulley is a hollow cylinder with a mass ofM = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away.
___________________m/s
(b) Find the angular speed of the pulley at the same moment.
______________rad/s
Explanation / Answer
m1 - .495 from .395
m2 = .795 from .840
M .35 from .350
r1=.02 = .02
r2=.03 -.03
mk = .250 =.250
vi=.820m/s = .82
(a)
I(cylinder) = 0.5*0.35*(0.02^2+0.03^2) =2.275*10^-4
By conservation of energy:
initial KE = 0.5*0.795*0.82^2+0.5*0.495*0.82^2-0.25*0.84*9.8*0.7+0.5*2.275*10^-4*(0.82/0.03)^2
= -0.9219
Final KE = 0.5*0.795*V^2+0.5*0.495*V^2-0.495*9.8*0.7+0.5*2.275*10^-4*(V/0.03)^2
=-3.3957+0.771389 V^2
Thus:
0.74389V^2-2.7097 = -0.9219
V=1.791
(b)
w = 1.791/ sqrt(.02^2 + .03^2)
=49.6734 rad/s