Tension = mg-ma = 4.68278N
Torque = (mg-ma) x r = 1.404834Nm
Iα = 0.199751046Nm this is acceleration Torque (required to create α)?
so fricitonal Torque=Torque-acceleration Torque (Iα)
i calculate 1.40483Nm-0.199751046Nm= 1.205078954Nm
frictional torque = 1.205078954Nm
i hope...
thank you for being so patient i am sure i have made this much more complicated than it really is
so Iα calculates the torque required to accelerate the flywheel to α ?
by total torque i mean the force of the mass (mg) x r = 0.5kg x 9.81ms^-2 x 0.3m =
1.4715Nm that would be the torque on...
sorry for the silence i have been away
but hopefully:blushing:i have got this one now
right so to calculate the acelerational torque i use I x Alpha
I=mk^2 = 3 x .212^2 x 1.481481481 = 0.199751046Nm
this is the torque required to accelerate the flywheel to the 1.481481481rads^-2 ...
the torque as a result of the Mass would be 4.905N x 0.3m =1.4715Nm
the frictional torque (is this the correct term?) is 4.682778N x 0.3m = 1.4048334Nm
so the net torque would be = 0.2222222N x 0.3m = 0.0666666Nm
my answer to the D therefore would be that the frictional torque (resisting...
for D
if the mass falls @ 9.81ms^-2 there is no resisting force.
the mass actually falls @ 0.4444ms^-2
so using I x alpha i calculate the acclerating torque this is the torque required to accelerate the flywheel?
but this is where i start to get confused i know the linear force that...
Homework Statement
i am not looking for the answers here just someone to steer me back on track as i feel i am moving away from the correct path.
i have answered some of the problem and i am a little unsure if i have tackled it in the correct way.
a mass of 0.5kg is suspended from a...