In the figure below, the hanging object has a mass of m 1 = 0.435 kg; the slidin
ID: 1477891 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.435 kg; the sliding block has a mass of m2 = 0.880 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 ofvi = 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.
(b) Find the angular speed of the pulley at the same moment.
rad/s
Explanation / Answer
I = ½m(R2² + R1²) = ½ * 0.350kg * (0.030² + 0.020²)m² = 2.275*10-4 kg·m²
For a non-slipping cord, = v/r, so the KE of the pulley is
KEp = ½I² = ½I(v/r)² = ½(I/r²)v² where v is the speed of the blocks and r = R2.
-PE = KE + work done
At the first reference point, both blocks have moved "x," so
-m1*g*-x = ½(m1 + m2 + I/r²)(vi)² + µ*m2*g*x
(m1 - µ*m2)g*x = ½(m1 + m2 + I/r²)(vi)²
x = (m1 + m2 + I/r²)(vi)² / (2g(m1 - µ*m2))
Plug in m1, m2, I (calculated above), r = R2, vi and µ and find
x = (0.435kg + 0.880kg + (2.275*10-4 kg·m² / 0.0302))*0.8202 / (2*9.8(0.435 - 0.250*0.880kg))
x = 0.250 m
"0.700 m away" means that x 0.950 m. Rearranging the above equation yields
v² = (2g(m1 - µ*m2))*x / (m1 + m2 + I/r²)
Plug in m1, m2, I, r, x and µ
v2 = 2*9.8(0.435 - (0.250*0.880))*0.950 / (0.435 + 0.880 + (2.275*10-4 / 0.2502))
v = 1.74m/s