Fake gravity

The well known formula for centripetal acceleration is $a_c=\frac{v^2}{R}$ as we (hopefully) all know. We also know that (tangential velocity) $v=\omega R$ , where $\omega$ is angular velocity.

The centripetal acceleration we want is g, since we are looking to simulate the acceleration of gravity on earth. Thus we have $g=\omega^2 R$ . We know $g$ , we know $R$ , so $\omega=\sqrt{\frac{g}{R}}$ . Plugging in the numbers, we find that $\boxed{\omega=0.313 rad/s}$ .

Working in an inertial frame, force that the rim exerts on the person $($ which is the normal reaction $)$ equals the force that causes the circular motion i.e.

$N=\:$ $\displaystyle{F_{centripetal}}=\:$ $Force\;felt\;on\;earth\;due\;to\;gravity$ $\big($ As given in the question $\big)$

$\implies\:$ $m$ $r$ $\omega^{2}=\:$ $m$ $g$

$\implies\:$ $\omega=\:$ $\displaystyle{\sqrt{\frac{g}{r}}}$

Plugging in $r=\:100\;m$ and $g=\:9.8\;m\,s^{-2}$ we find that $\omega=\:$ $0.313\;rad\,s^{-1}$ which is what we seek.

For the person on the cylinder to experience free fall acceleration, the centripetal acceleration must be equal to 9.8m/ $s^{2}$ . $\frac{v^{2}} {r}$ =9.8m/ $s^{2}$ ,and v=w*r. solving for w gives 0.313rad/s

If a person of mass $m$ is standing on the rim of the spaceship which has the shape of a cylinder , the centrifugal force force experienced by him in the rotating spaceship is $F_{c}$ = $m r ω^2$ . If $F_{c}$ is equal to the force of gravity felt by the person on Earth , then $F_{c}$ = $mg$ . Thus , $m r ω^2$ = $mg$ which implies that $ω$ = $\sqrt\frac{g}{r}$ . Substituting the values of $r$ and $g$ , we get $ω$ = $0.313 rad/s$

The cilinder must spin with a centripetal acceleration equal to the acceleration due to gravity. That is $a = \omega^2 R = g$ . So, we solve for $\omega = \sqrt{\frac{g}{R}} = \sqrt{\frac{9.8}{100}} = 0.313$ rad/s

If the force felt on the rim of the cylinder matches the force of gravity felt on earth, the centripetal force is the same as the force of gravity. Therefore, we have $F_c=F_g$ , or that $\frac{mv^2}{r}=mg$ . The two $m$ cancel out, and $v=\omega r$ , so we have $\omega^2r=g$ . Plugging in our values for $r$ and $g$ , we can solve for $\omega$ , getting that $\omega = \sqrt{\frac{9.8}{100}} \approx \fbox{0.313}$ .

Yo,this is so tricky,

to get the same equal force,

ac=ag=9.81m/s/s, as v=rw, ac=ag=v^2 /r, w= ( 9.81/100) ^ 1/2 = 0.313 rad/s...thanks

g = r*w^2 Hence w^2 = (9.8/100) so that, w = 0.313 rad/s

In order for the force felt by a person on the cylinder to match the force of gravity, the centripetal force must be equal to the value of the force of gravity. We can also neglect mass, because the mass of an object in space is equal to the mass of the object on Earth. So, the centripetal acceleration must be equal to the acceleration due to gravity.

The centripetal acceleration of an object is given by

$a = {v^2}/r$

(a is centripetal acceleration, r is radius, and v is speed of the object at the end of the given radius.)

Substituting in $r=100$ m and $a = g = -9.8$ m/s^2 and solving for $v$ , we find $v=\sqrt{980}$ m/s

We have to convert this into rad/s. multiply $v=\sqrt{980}$ m/s by 1/100 radians per meter (r=100 m) to get v in rad/s as 0.313

Velocity(v) = $r\omega$ and Acceleration(a) = $v^2/r$

Therefore, $a=r\omega^2$

$a/r=\omega^2$

$\sqrt{a/r}=\omega$

Subsituting , $a=9.8$ and $r= 100$ We get, $\omega=0.313$ rad/s

N = m.w².r and N = m.g

m.w².r = m.g

w².r = g

w² = g/r

w² = 9,8 / 100 ---> w² = 0,098 , then w = 0,313 rad/s

Since we wish to simulate a gravitational effect, the centripet force must be equal to the weight (mg). So: Fc=mg ---> mw^2(100)=m(9.8) --> w~3.13

In order for the force to feel the same at the rim of the cylinder as it would on the top of Earth, the centripetal acceleration must be equivalent to gravitational acceleration on Earth. This means $r\omega^2 = g$ Solving for angular velocity, we obtain $\omega = \sqrt{\dfrac{g}{r}} = \sqrt{0.098} = 0.313 \dfrac{rad}{s}$ which is our final answer.

ma=mwr^2

w=squareroot(a/r)

w=squareroot(9.8/100)

Therefore w=0.313

Being:

$a_c$ = aceleración centrípeta

$w$ = velocidad angular

$r$ = radio del cilindro

$a_c = w^2 \times r$

so:

$9.8 = w^2 \times 100$

$w = 0.313 rad/s$

ma=mω²R, ω=sqrt(a/R) = =sqrt(g/R) = =sqrt(9.8/100)=0.313 rad/s

Here , the force caused by the acceleration due to the gravity on any object having mass of 'm' will be mg & equal to centrifugal force mw^2r. so, g=w^2r . Hence, w=.313

w= √(g/r) = √(9.8/100) = 0.313

Since the space ship is spinning around its axis the force felt by the astronaut standing on the rim of the cylinder is the centripetal force, $mω^{2}R$ . The astronaut experiences the gravity force felt on earth if the centripetal force is equal to his weight, $mg$ . So we have $mω^{2}R=mg$ and we find that $ω=\sqrt{g/R}$ .

Here, since the cylinder is rotating its has some centrifugal force acting on the objects within it and this can be compared to the the force due to gravity. It is known that Centrifugal force is opposite to the direction of the centripetal force.

Thus, $F_{centrifugal} = - m \omega ^{2} R$

where R is the radius of the cylinder.

$\therefore mg = F_{centrifugal}$

i.e $mg = - m \omega ^{2} R$ where $\omega$ is the angular velocity.

$\Rightarrow \omega = \sqrt { \frac{-g}{R}} = \sqrt { \frac{9.8}{100}} = 0.313$ rad/s

Thus, angular velocity is equal to 0.313 rad/s

Force felt on Earth = Weight = $mg$
where m is mass of person, & g is acceleration due to gravity on earth .

Force acting on the body in the spinning space ship = Centrifugal force = $mω^{2}r$ where ω is the angular velocity and r is the radius of the spaceship .

But it is given that they are equal . Thus,

$mg = mω^{2}r$

$g = ω^{2}r$ as mass is non-zero

$ω = \sqrt{g/r}$

Putting all values in S.I. units , we get

$ω = 0.313 rad/s$

If the gravity is equal in the space ship, we have:

P=Fcp

mg=mw $^{2}$ R

w= $\sqrt{0.098}$

w=0.313

Let

$\mathrm{r}$ = Radius of the spaceship

$\mathrm{m}$ = Mass of the astronaut and

$\omega$ = Angular velocity (all in SI units)

From, Newtons $2^{nd}$ Law,

$\mathrm{mr\omega^2}$ = $\mathrm{mg}$ where $\mathrm{g}$ is the acceleration due to gravity of earth

On solving the equation, we get

$\omega$ = 0.313 $\mathrm{rad/s^2}$

$a = \frac{v^{2}}{r}$

$v = \omega r$

$\omega=\sqrt{\frac{g}{r}}$

Firstly, lets define some variables.

$\mathbf{r}=$ the displacement of the astronaut's feet on the edge of the ship from the axis of rotation.

$|\mathbf{r}|=100m$

$\omega=$ the angular velocity of the ship, the value to be determined, but is constant as time progresses.

$v=$ the velocity of the astronaut's feet on the edge of the ship $=\mathbf{r}\omega$ .

$a=$ the acceleration of the astronaut's feet on the edge of the ship. Because we want to simulate gravity, $a=9.8ms^{-1}$

$d\mathbf{r}=$ the small change in $\mathbf{r}$ caused by the astronaut's feet having velocity $v$ .

If you are struggling to visualise what $d\mathbf{r}$ looks like, draw a circle, and then draw in the vector $\mathbf{r}$ . Then draw a vector from the centre of the circle to the circumference, just to the side of where $\mathbf{r}$ meets the circumference. Call this new vector $\mathbf{r}+d\mathbf{r}$ . Then draw a vector from the head of $\mathbf{r}$ to the head of $\mathbf{r}+d\mathbf{r}$ . You can see that this vector is $d\mathbf{r}$ , the small change in $\mathbf{r}$ caused by the astronaut's feet having velocity $v$ .

We also know that $a=\frac{dv}{dt}=\frac{d(\mathbf{r}\omega)}{dt}$ .

Using the prooduct rule for differentiation:

$a=\mathbf{r}\frac{d\omega}{dt}+\omega\frac{d\mathbf{r}}{dt}$

Because $\omega$ is constant as time progresses, $\frac{d\omega}{dt}=0$ . This means that:

$a=\omega\frac{d\mathbf{r}}{dt}$

We also know that $\frac{d\mathbf{r}}{dt}=v$ , meaning that $a=\omega v$ .

Plugging in $v=r\omega$ gives $a=\omega^2r$

Rearranging gives $\omega=\sqrt{\frac{a}{r}}$ , and plugging in the numerical values gives:

$\omega=\sqrt{\frac{9.8}{100}}=\frac{7\sqrt{5}}{50}=0.313rad \text{ }s^{-1}$ to three significant figures.

The spinning produces centripetal force, which is then used to create some force so that a person standing on the rim of the cylinder feels as if the gravity force works. Assuming the mass of the person is m, we get:

$mg=ma_{cp}$

$mg=mw^2r$

$g=w^2r$

$w=\sqrt{g/r}$

$w=\sqrt{9.8/100}$

$w=0.313$

$ma = mω^{2}R$

$ω = sqrt{\frac{a}{R}}$

= $\sqrt{\frac{9.8}{100}}$

= 0.3130495168499706

At equilibrium:- (Weight of the body) ma=mω²R, (Centripetal force)

ω=sqrt(a/R) =sqrt(g/R) =sqrt(9.8/100)=0.313 rad/s

Let the mas of ship on earth be $m$ . So the force $F_g$ on earth will be $m.g$ Now as the ship is in space and it experiences a centripetal force $F_c$ equal to $m\omega^{2}r$ . These forces that are $F_c$ and $F_g$ must balance each other. So we have,

$F_c$ $=$ $F_g$

$m\omega^{2}r$ $=$ $m.g$

$m$ cancelling each side so,

$\omega^{2}r$ $=$ $g$

$\omega^{2}.100$ $=$ $9.8 m/s^{2}$

Solving we get $\omega$ $=$ $0.313$

Use F = m x ω x r² where F is N + m x g. So N + m x g = m x ω x r². There is no normal Force because he is at space . so we get ω = √g/r = 0.3130 rad/s

$F=m g=m r \omega^2\\ 9.8=100 \omega ^2\\ \omega=0.31305$

Since the force felt by a person standing on the rim of the cylinder matches the force of gravity felt on Earth, the centripetal acceleration $a_r$ must equal the acceleration of gravity on earth. We have $a_r = r \omega^2$ , where $r$ is the radius and $\omega$ is the angular velocity. Then $\omega = \sqrt{\frac{a_r}{r}} = \sqrt{\frac{9.8}{100}} \approx 0.313.$

let the force be R, R = m w^2 r' also, R = mg combining, we get, w = (g/r)^0.5 = 0.313

When the cylinder rotates, the only force acting on the person is centrifugal force (assuming the spacecraft doesn't accelerate). For the force felt by the person to match the gravity on the Earth, this centrifugal force has to match with the gravitational force. Let $m$ be the mass of the person, let $v$ be the (linear) velocity of the rim of the cylinder, and let $r$ be the radius. Then, we have $\frac{mv^2}{r} = mg \implies v= \sqrt{gr}$ . The angular velocity will be then $\frac{v}{r} = \sqrt{g/r}$ . Plugging the values gives the result $0.313 rad/s$ .

By using Centripetal Force Law,

$m\omega^2R = mg$

$\omega = \sqrt{\frac{g}{R}}$

By substituting $g = 9.81$ and $R = 100$ hence $\omega = 0.313$ rad/s

Since the formula for the magnitude of centrifugal acceleration due to spinning at constant angular velocity is (w^2) r we simply wright r=100 and solve (w^2) r=g.

In this case, we have $Fcen = mg$ . Therefore, $w^2r = g$ and $w = \sqrt{\frac{g}{r}}$ . With the constants given, we have: $w=0.313 rad/s$

$r {\omega}^2 = g$

Therefore $\omega = \sqrt {\frac {g}{r}}$