3 votes

Why can’t we accelerate a rotating object after some point?

This gets back to the original answer by Tigerguy. You have the maximum torque your legs can put out. You also have the maximum power your legs can put out (power = torque x rotational speed). But you ...
PatMc's user avatar
  • 109
3 votes

Effect of Earth's rotation on moving vehicles

The answer to the effects of Earth's rotation of moving vehicles: First, Earth doesn't rotate at a rate of 460 m/s. In fact, m/s is not the unit for the angular speed. It's like using Liters for mass. ...
Sina Atalay's user avatar
2 votes

How does 2nd gear create more thrust than 1st gear?

I think that you've totally missed the limiting factor - maximum engine speed. The 1st gear has a maximum velocity value after which you can’t accelerate more. No, it's the engine RPM is limited. ...
Transistor's user avatar
  • 10.6k
2 votes

Why can’t we accelerate a rotating object after some point?

This would work if input power could handle infinite speeds. On a bicycle, your legs limit how fast you can pedal and the power you can input to the system. Motors have speed limits as well. The ...
Tiger Guy's user avatar
  • 6,803
2 votes

Changing the relative equilibrium position for a 2-mass spring-mass system

At equilibrium $\ddot{r}_2=0,\ \dot{r}_2=0$. So, from the second equation, $$r_2-r_1 = \frac{m_2 g \cos\theta}{K_{12}}$$ For positive values of $m, g, \cos\theta, K_{12}$, $r_2 > r_1$. i.e. mass #2 ...
AJN's user avatar
  • 1,039
2 votes

Calculating velocity of two different weight in a system with spring

The total energy of the system is $T=4 g\ \frac{2(200)}{1000}+2\ (3 g)\ \frac{ 500}{1000}=\frac{23 \ g}{5}=\frac{1127}{25}$ At a value $\theta$, the position of the $4 \ kg $ mass is $\{p_{x1},p_{y1}\}...
Suba Thomas's user avatar
  • 1,971
1 vote

Keeping a cardholder hanging from a lanyard always facing the same side

If the cardholder can be attached to the lanyard at both corners rather than at one point in the middle then it will tend to straighten out. Alternatively a piece of self-adhesive Velcro (hook side) ...
Transistor's user avatar
  • 10.6k
1 vote

Jitter Experienced by Spacecraft from Reaction Wheel Imbalances

In the real system you can assume that the presence of reaction wheels means that you can model the angular velocity and pointing angle as being the intended value, plus errors. If you're only ...
TimWescott's user avatar
  • 2,522
1 vote

Distributed load on a slope/axle

Net force and torque I assume $T_1$ and $T_2$ denote tension forces in the belt. From the outside perspective, only $T_1$ and $T_2$ and their respective directions should be relevant. They will ...
Tomáš Létal's user avatar
1 vote

Distributed load on a slope/axle

If you want to calculate the resultant force due to a variable distrubuted load on a generic profile, you can follow this line of reasoning (if you're not interested in the reasoning, scroll to the ...
RocketScience's user avatar
1 vote

How to calculate the centripetal force along a cartesian axis?

let's say your centripetal vector is rotating with an angular speed of $$\omega= \alpha t $$ $\alpha =$ angular acceleration Let's say the instantaneous x and y of the vector on a frame aligned with ...
kamran's user avatar
  • 22.2k
1 vote

How does 2nd gear create more thrust than 1st gear?

The torque an engine can produce depends on the RPM of the engine. The torque it can produce actually decreases after a certain point. See e.g. the graphs on the bottom of this page: While ...
Chris_abc's user avatar
  • 759

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