- #1
poortech
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1. A mass of 0.5 kg suspended from a flywheel, if the mass is released from rest and falls a distance of .5m in 1.5s calculate
a) the linear acceleration of the mass
b) the angular acceleration of the wheel
c) the tension in the rope
d) the frictional torque, resisting motion
mass of wheel = 3kg
outside radius of wheel = 300mm
radius of gyration = 212mm
2. Homework Equations
I = Mk^2
K.E. = 1/2 Iw^2
θ = 1/2at^2
T = work done/angular desplacement
T= m(a+g)
3. The Attempt at a Solution
a)
v = .5/1.5 = .33m/s
a = v-u/t
.33/1.5 = .22m/s^2
b) angular acceleration = a/r
0.22/0.3 = .73m/s^2
c) T = m(a+g)
0.5(.22+9.8) = 5.01 N
d)Frictional Torque
moment of interia I = mk^2
0.3 x (.212)^2 = .135 kg.m^2
displacement θ = 1/2 at^2
1/2 (0.73)(1.5)^2 = .821 rad
K.E = 1/2 Iw^2
1/2 (.135) (1.1)^2 = .0816J
T = work done / angular displacement
.0816/0821 = .0993 Nm
Plz let me know if I am doing it right, specially Frictional Torque...
a) the linear acceleration of the mass
b) the angular acceleration of the wheel
c) the tension in the rope
d) the frictional torque, resisting motion
mass of wheel = 3kg
outside radius of wheel = 300mm
radius of gyration = 212mm
2. Homework Equations
I = Mk^2
K.E. = 1/2 Iw^2
θ = 1/2at^2
T = work done/angular desplacement
T= m(a+g)
3. The Attempt at a Solution
a)
v = .5/1.5 = .33m/s
a = v-u/t
.33/1.5 = .22m/s^2
b) angular acceleration = a/r
0.22/0.3 = .73m/s^2
c) T = m(a+g)
0.5(.22+9.8) = 5.01 N
d)Frictional Torque
moment of interia I = mk^2
0.3 x (.212)^2 = .135 kg.m^2
displacement θ = 1/2 at^2
1/2 (0.73)(1.5)^2 = .821 rad
K.E = 1/2 Iw^2
1/2 (.135) (1.1)^2 = .0816J
T = work done / angular displacement
.0816/0821 = .0993 Nm
Plz let me know if I am doing it right, specially Frictional Torque...