Question

A counterweight of mass m = 4.30 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 = 1.60 kg. The spokes have negligible mass. (a) What is the net torque on the system about the axle of the pulley? magnitude N · m direction (b) When the counterweight has a speed v, the pulley has an angular speed ω = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley. ( kg · m)v (c) Using your result from (b) and tau with arrow = dL with arrow/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) m/s2

Answer 1

given

m = 4.30 kg

R = 7.00 cm

M = 1.60 kg

a )

_total = _wheel + _{counterweight}  

= r X F

we have r = - R i and F = - T j

= ( - Ri ) X ( - T j )

= R T k ( since i X j = k )

b )

total angular momentum

L_total = L_wheel + L_{counterweight}  

we have L = I

now L_wheel = M R² => M R v

L_{counterweight}   = ( - Ri - Qj ) X ( - m v j )

L_{counterweight}   = m R v k

L_total = ( m + M ) R v k

c )

given

= dL / dt

the equation is used here is = m g R

dL /dt = ( m + M ) R dv / dt

m g R = ( m + M ) R a ( since a = dv/dt )

a = m g / ( m + M )

d )

given

m = 4.30 kg

R = 7.00 cm

M = 1.60 kg

using a = m g / ( m + M )

a = 4.3 X 9.8 / ( 4.3 + 1.6 )

a = 7.14 m/sec²