The 3.5 kg, 33-cm-diameter disk is spinning at 320 rpm.
2 Answers
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take it as an example——————
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Moment of inertia of disk = (MR^2)/2
= 2.2*0.2^2/2
= 0.044 Kg m^2
Angular acceleration a :
w = w’+ at
0 = [380*2*(22/7)]/60 + a*3
0 = 39.8 + a*3
a =13.27 rad/s^2
Torque = moment of inertia*angular acceleration
= .044*13.27
= 0.58 Nm
Torque = F*r
F = Torque/r
= 0.58/0.2
= 2.9 N
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angular momentum = Iω
I = inertia
ω = omega (rad/s)
320/9.55 = omega = 33.5 rad/s
I = 1/2mr^2 = 0.5 x 3.5 x 0.165^2 = 0.047644 kg-m^2
0.047644 x 33.5 = 1.596 kg-m^2/s
force acting = change in momentum/ time taken for change to happen.
0 – 1.596 = – 1.596 kg-m^2/s = change in momentum
-1.596/2.5 = – 0.6384 N
