7. -4 points SerPSET9 11.P.018 My Notes Ask Your Teacher A counterweight of mass

Question

7. -4 points SerPSET9 11.P.018 My Notes Ask Your Teacher A counterweight of mass m = 3.80 kg is attached to a light cord that is wound around a pulley as shown in the figure below. The pulley is a thin hoop of radius R = 7.00 cm and mass M = 1.80 kg. The spokes have negligible mass. (a) What is the net torque on the system about the axle of the pulley? magnitude direction Select- N·m (b) when the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley kg·m)v

Explanation / Answer

given counterweight m = 3.8 kg

R = 7 cm = 0.07 m

M = 1.8 kg ( mass of pulley)

a. Net torque = t*r ( where t is tension in the string)

let acceleration of the system be a

then

mg - t = ma

t = Ia/r

t = I*(g - t/m)/r

tr = Ig - It/m

t(r + I/m) = Ig

t = Ig/(r + I/m)

torque T = t*r = Igr/(r + I/m)

T = 0.02744*9.81*0.07/(0.07 + 0.02744/3.8)

T = 0.244 Nm

b. total angular momentum = I w

I = moment of inertial of the system

I = mR^2 + MR^2 = 0.02744 kg m^2

hence

angular momentum = 0.02744 v / 0.07 = 0.392 v kg m

c. T = dL/dt

but L = 0.392 v

hence

T = 0.392 dv/dt = 0.244

dv/dt = a = 0.622 m/s/s

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