In the figure below, the hanging object has a mass of m 1 = 0.495 kg; the slidin
ID: 1437613 • 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 = 0.845 kg; and the pulley is a hollow cylinder with a mass of M = 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
a)
area density, rho = M / pi (R2^2 - R1^2) = 222.82 kg /m^2
moment of inertia of pulley, I = [rho * pi R2^2 *R2^2 /2 ] - [ rho* pi R1^2 * R1^2 /2 ]
= pi*rho [ R2^3 -R1^3] / 2
= 6.65 x 10^-3 kg m^2
friction force = uk m2g = 0.250 x 0.845 x 9.8 = 2.07 N
Using work energy theorem.
work done by friction + work done by gravity = change in KE
- 2.07x 0.7 + (0.495 x 9.8 x 0.7) = (0.495 + 0.845) ( v^2 - 0.820^2 ) /2 + I(wf^2 - wi^2)/2
1.95 = 0.67( v^2 - 0.82^2) + (0.00665 ((v/0.03)^2 - (0.82/0.03)^2 ) /2 )
1.95 = 0.67v^2 - 0.45 + 3.69 v^2 - 2.48
v = 1.06 m/s
b) w = v/R2 = 1.06 / 0.03 =35.3 rad/s
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