In the figure below, the hanging object has a mass of m1 0.480 kg; the sliding b

Question

In the figure below, the hanging object has a mass of m1 0.480 kg; the sliding block has a mass of m2 0.840 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 vi0.820 m/s toward the pulley when it passes a reference point on the table. RI mo mi (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

mass of hanging object = m1 = 0.48 kg

mass m2 = 0.84 kg

M = 0.35 kg ( hollow cylinder)

R1 = 0.02 m

R2 = 0.03 m

coefficient of kinetic friction k = 0.25

vi = 0.82 m/s

a. let thickness of pulley be t

then densioty of pulley, rho = M/pi(Ro^2 - Ri^2)t

moment of inertial of pulley = I = 0.5(Ro^4 - Ri^4)rho*pi*t

I = 0.5(Ro^4 - Ri^4)M*pi*t/pi(Ro^2 - Ri)^2*t

I = 0.5M(Ro^2 + Ri^2) = 0.5*0.35*(0.03^2 + 0.02^2) = 0.0002275 kg m^2

also, initial angular speed of pulley wi = v/R2 = 27.333 rad/s

hence total energy, iitially, Ei = 0.5m1vi^2 + 0.5m2vi^2 + 0.5*I*wi^2

Ei = 0.52876581612375 J

gain in kinetic energy due to falling of mass m1 by d = 0.7 m

Eg = m1*g*d = 3.29616 J

loss in energy due to friciotn

Ef = 0.25*0.84*9.81*0.7 = 1.44207 J

hence final total KE = Ef

Ef = Ei + Eg - Ef = 2.3828558 J

let final speed be v

then

0.5(m1 + m2)v^2 + 0.5I(v/R2)^2 = 2.3828558

v = 1.7404 m/s

angular speed = w = v/Ro = 58.0241 rad/s

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