If I may interject…
I think that in Post #59, @kuruman is considering the case where the vector C and the axes are all rotated through the same angle.
But you are considering the case where C is fixed and the axes are rotated (i.e rotation of coordinate system only).
In the former case, the...
I assume your attachment is referring to ‘case 2’, with the mass on the right of the equilibrium point.
Your eqation ##m \ddot x = -kx\hat i+ 2b \hat i## is only ‘a snapshot’. The (variable) velocity ##\dot x## has disappeared because you have replaced it by a specific value (-2m/s).
(By the...
Yes, but note that the restoring force is ##-kx\hat i## in all 4 cases.
No. The damping force is always ##-bv\hat i##.
In case 2 the mass is moving left, so ##v## is negative. The damping force is:
##\vec F_{damping} = -bv\hat i##
##= - \text {(a positive quantity} \times \text {(a negative...
Maybe this will help.
F = -kx – bv where k, b >0.
This is valid at all positions. We can check-out all the possibilities...
Call the equilibrium position ‘O’. There are 4 cases to consider:
1. The mass is to the right of O, moving right and slowing down. In this case:
x is positive, so kx...
There are a couple of mistakes I can spot.
For the part 1 journeys, the cyclists leave A and B simultaneously, meeting 2km from B. But do the cyclists arrive at A and B simultaneously? So (if neither cyclist waits) would they start their part 2 (return) journeys simultaneously? Does this...
For information, the average value of ##sin(x)## over a positive half-cycle is ##\frac 2{\pi}##.
With ##\mathscr E =NBA\omega \sin (\omega t)## and ##\omega = \frac {2 \pi}T##, this gives the OP's post #1 formula for the average value of a fully rectified output.
The formula is (I believe) correct when finding:
a) the average emf of a simple DC generator, or
b) the average emf for the fully-rectified output of a simple AC generator.
If you know a little calculus, find the average value of ##\sin x ## over, say, the positive half of a cycle; then the...
If you raise a number to some power, the power must be dimensionless (a pure number) So raising something to the power ##M##, where ##M## is mass, is wrong.
The question has other issues as well IMO. It would be interesting to know where the original question comes from.
Probably best to avoid the use of the word 'potential' as it has specific meanings in physics. In fact '(ma)d' equals the work done by the net force (if I understand you correctly).
If a fixed force is applied for a fixed time (i.e. a given impulse) then I think 'd ' must increase - so that...
Many thanks. Eyes (and brain) going downhill. Have edited.
I understand your difficulty. Have you ever played snooker/billiards/pool? The cue ball moves in the direction of the cue’s motion – even if you hit the ball off-centre, giving it spin.
The force from the cue (for an off-centre...
Not sure where you mean - can't see anything wrong. But am happy to make a correction if there's a mistake.
The friction is simultaneously doing two jobs - changing the linear momentum and (because it provides a torque) changing the angular momentum. This is what actually happens! The same...
Use a free body diagram (FBD). Always!
During the rolling-and-slipping phase, the torque is provided by the (full) kinetic frictional force - as you can see on the FBD. So you can easily find the torque and angular acceleration .
During the rolling-and-slipping phase, the tranlational...
Can I add this to the discussion.
With an energy-based approach here, work done by kinetic (sliding) friction is a combination of the heat generated, the change in linear KE, the change in rotational KE. This ‘3-way split’ makes it messy.
The problem can be solved quite easily using...