Can the OP or someone else explain how that equation was simplified to get a factor of 3 in the denominator for "m"?
The original equation involves a term with rotational kinetic energy (1/2Iω2).
Was that "I" for the object's inertia converted to some kind of equivalent mass?
Interesting, so basically what we are interested in here is fluid friction and rolling friction.
If a car is traveling at constant speed, and we somehow managed to suddenly remove fluid friction and rolling friction - there would be a net torque at the wheels and hence it would begin to...
Thank you for your response.
Yes, you are right...I should have considered the direction of the vectors. Centripetal force is pulling inwards and that is not aligned with torque.
I do not understand how there can be no torque at the wheel when it is moving at a constant speed though.
Sure, we...
The way I have understood it is that dynamic torque involves angular acceleration, while static torque has zero angular acceleration:
α =Δω/Δt = 0
But static torque still has centripetal acceleration with constant speed:
ac = v2/r
where v is linear velocity, and r is the radius
Then F = mac...
Hi SteamKing,
I am not saying it is a problem, nor am I looking to put a damper on it. But for the sake of modelling it precisely it has a torsional viscous coefficient parameter. For the most part, it is very very small, usually 10^-4 or even smaller.
In any case, I need to put the...
I have some data from an electric motor found here:
http://www.engelantriebe.de/pdf/DAT_HLR26_11-13_engl.pdf
However, it does not include information about the coefficient of torsional viscous damping which I need for a Simulink model.
The units for torsional viscous damping are Nm.s/rad...
Homework Statement
In the equation dm = δ x 2∏rLdr
Where δ = density, and
2∏rL = volume
How is it that the volume can be 2∏rL? The units of r is (metres) and the units of L is (metres) which leads to m2 (Area)
Should it not be ∏r2L for volume?
The Attempt at a Solution...
But you still need to take into account the frictional force.
What I did was:
F = mg = 8829N
Fn (normal force) = cos 30 x 8829 = 7646N
Ff (forward force down slope) = sin 30 x 8829 = 4414.5N
Then the frictional force is: 0.3 x 4414.5N = 1324.35N
Fr (resultant force) = 4414.5N -...
Leaving aside some of the interesting technicalities of the question, would the working out be correct? I kind of did it differently and come to a slightly different answer and I just want to make sure the solution itself is correct.
I am not sure where exactly the question came, it was part of our tutorial questions -- I would assume the lecturer made it up. It's not really a straight out "physics" course as such, it's about electric motors.
I don't think he's trying to say the coefficient of friction is enough to stop it...
Homework Statement
Homework Equations
A few basic conversions/equations:
50km/h = 13.89m/s
ω = 13.89/0.3 = 46.3 rad/s
Torque = Force x Radius
Power = Torque x Angular Velocity
The Attempt at a Solution
Solution is already given above in the picture, but I am confused by...
Hi guys,
I'm wondering whether I'd be able to generate around 3V - 5V with a magnetic coil using a high turn ratio in the presence of a changing magnetic field.
I know guitar coils can generally produce around 500mV - 1V, I wish to generate around 3V -5V even if it is for a very short time...