Question

In the figure below, the hanging object has a mass of m₁ = 0.370 kg; the sliding block has a mass of m₂ = 0.900 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 μ_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 v_i = 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.

Answer 1

I = ½m(R2² + R1²) = ½ * 0.350kg * (0.030² + 0.020²)m² = 2.275e-4 kg·m²

For a non-slipping cord, ω = v/r, so the KE of the pulley is

KEp = ½Iω² = ½I(v/r)² = ½(I/r²)v² where v is the speed of the blocks and r = R2.

-ΔPE = ΔKE + 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 = 0.36 m

"0.700 m away" means that x ≈ 0.700+0.36 = 1.06 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 and find

v = 1.406 m/s

b)w = v/r = 1.406/0.03

= 46.88 rad/s