Weight required to hang straight down with known torque

[MoMan]

How much Weight (W) is required to keep a weight hanging straight down when the torque is known on a rotating shaft? Please see attached image. Assume a lever weight of zero to keep it simple.

 

[berkeman]

it simple

Yeah, $\infty$

 

[jbriggs444]

How much Weight (W) is required to keep a weight hanging straight down when the torque is known on a rotating shaft? Please see attached image. Assume a shaft weight of zero to keep it simple.View attachment 303451

Suppose that the shaft is deflected so that it does not hang straight down. Suppose that it is deflected by $x$ meters rightward, for instance. Do you know how to calculate how much restoring torque results from gravity pulling on the deflected weight?

There is a simple answer to the question you ask. But the Physics Forums way is to guide you into discovering that answer for yourself.

 

[hmmm27]

What's the application ?

 

[Delta2]

The way to answer this question with infinities: The torque of weight around the central point of the shaft is $T_W=W\cdot 1m\cdot \sin 0$. For any finite W this equals to zero and hence the total torque will be equal to the torque from shaft and it will rotate the weight.
However if $W=\infty$ then $$T_W=\infty\cdot 0=\text {maybe something finite and equal to -}T_{shaft}$$ and hence for infinite weight the system might not rotate.

 

[berkeman]

What's the application ?

It's either a puzzle (and not a very good one, IMO), or he's just trolling.

 

[erobz]

At what point do macro scale accelerations become imperceptible? Can we not effectively consider the rotation of $0.1 ^{\circ}$ over 10 years as effectively no acceleration. We do this kind of thing all the time in our modeling.

 

[jbriggs444]

At what point do macro scale accelerations become imperceptible? Can we not effectively consider the rotation of $0.1 ^{\circ}$ over 10 years as effectively no acceleration. We do this kind of thing all the time in our modeling.

Well after the point at which suspended weights break the mechanism.

 

[hmmm27]

$$m=\frac\tau{\mu_s{rg}}$$