# Rotational velocity necessary to make apparent weight equal to “real” weight?

I have been trying to find a solution to this, but I never reach same one that my professor provides. The problem is the following:

For an object of mass m standing still on the surface of the Earth, find what the planet's rotational speed ω should be so that the apparent weight is equal to the "real" weight (W=mg). The trivial solution (ω = 0) isn't valid.

This comes from a chapter about non inertial reference frames. The exercise doesn't mention anything about latitudes.

Since I know that the apparent force exerted on a body is equal to the real forces minus the fictitious forces, all I can come up with is: $$F_{apparent} = F_{real}-F_{fictitious} = W + N - (F_{centrifugal})$$

(W being weight, N being the normal force, and $$F_{centrifugal}$$ being the centrifugal force). There is no Coriolis force since the object isn't moving with respect to the Earth's surface. By setting that the apparent foce should be equal to the weight, I get: $$F_{apparent}=-mg\hat{r}+mg\hat{r}+m\omega^2r\hat{r} \Rightarrow -mg\hat{r} = -mg\hat{r}+mg\hat{r}+m\omega^2r\hat{r}$$ Therefore, $$\omega=\sqrt{g/r}$$

But the solution given by my professor is $$\omega = \sqrt{2g/R_{Earth}}$$, which is $$\sqrt{2}$$ times my solution. Where did I go wrong?

• A little more context would be helpful. – David White May 24 '19 at 0:15

$$\vec{F}_{real} = -mg\hat{r}$$ in spherical coordinates, i.e. this force is towards the center of mass of Earth, not the axis of rotation.

$$\vec{F}_{apparent} = \vec{F}_{real} - \vec{F}_{fictitious}$$ $$\vec{F}_{fictitious} = \vec{F}_{centifugal} = +m\omega^2\rho\hat{\rho}$$ for $$\hat{\rho}$$ is the vector from the axis of rotation to the point of interest. We can seperate $$\hat{r}$$ into cylindrical coordinates as follows:

$$\hat{r} = \hat{z}\cos(\theta)+\hat{\rho}\sin(\theta)$$ Inserting all back into the second equation, $$\vec{F}_{apparent} = \vec{F}_{real} - \vec{F}_{fictitious}$$ $$\vec{F}_{apparent} = mg(\hat{z}\cos(\theta)+\hat{\rho}\sin(\theta)) - m\omega^2\rho\hat{\rho}$$

Now, obviously, $$\hat{F}_{apparent} \neq \hat{F}_{real} = \hat{r}$$ for nonzero $$\omega$$, therefore I guess what he was asking for was a solution for which the magnitudes of these two are equal: $$|\vec{F}_{apparent}| = \sqrt{(mg\cos(\theta))^2+(mg\sin(\theta) - m\omega^2\rho)^2}$$ thanks to the fact that $$\hat{\rho} \perp \hat{z}$$. Then, $$|\vec{F}_{apparent}| = \sqrt{(mg\cos(\theta))^2+(mg\sin(\theta) - m\omega^2\rho)^2} = |\vec{F}_{real}| = mg$$ $$g^2(\cos(\theta))^2 + g^2(\sin(\theta))^2 - 2g\omega^2\rho\sin(\theta) + (\omega^2\rho)^2 = g^2$$ $$g^2 - 2g\omega^2\rho\sin(\theta) + \omega^4\rho^2 = g^2$$ $$\omega^4\rho^2 = 2g\omega^2\rho\sin(\theta)$$ $$\omega=0$$ or $$\rho=0$$ (both trivial) or $$\omega^2\rho = 2g\sin(\theta)$$

Then $$\omega = \sqrt{\frac{2g\sin(\theta)}{\rho}}$$ Since we know $$\rho = R_{Earth}\sin(\theta)$$, $$\omega = \sqrt{\frac{2g}{R_{Earth}}}$$

• I liked the rest of the answers, but I'll have to choose this one because it's the most comprehensive and mathematically accurate. Thanks for your time and dedication :) – mar May 24 '19 at 8:24
• I do have a question. I redid the exercise after reading your answer, but since $\hat{F}_{real}=-mg\hat{r}$, after replacing the vector $\hat{r}$ with $\hat{z}\cos\theta + \hat{\rho}\sin\theta$, I get a negative sign in the real force (which you did not), and I end up having a minus sign in the square root. Why is this? – mar May 24 '19 at 9:45
• Wow. Awesome answer. The only way to up it would be to factor in the ellipticity of the Earth and gravitational variance with latitude. – Paul Childs May 24 '19 at 10:03
• Yes, there is a typo in the signs, but they are swapped for both the gravitational and centrifugal forces. – Paul Childs May 24 '19 at 10:08
• @PaulChilds Can you suggest an edit? – acarturk May 24 '19 at 12:45

You should not have the normal force in there. The question is after apparent weight not force (else you'd apparently be weightless for no rotation). You also have the sign wrong. The trivial solution of zero rotation will have the apparent weight the same sign as the actual weight. For the rotating case it will have to be opposite, i.e.:

$$+mg = -mg+ m \omega^2r$$

The reading on bathroom scales is equal to the magnitude of the apparent weight of an object which has been placed on the scales.

If a mass $$m$$ is placed on such scale which are on the ground at the Equator of the planet, the magnitude of the force on the mass due to the scales $$X$$, the apparent weight of the mass, can be found by applying Newton’s second law

$$mg -X =mr\omega^2$$

where $$g$$ is the gravitational field strength, $$\omega$$ is the angular speed of the planet and $$r$$ is the radius of the planet.

This gives the trivial solution to the problem $$X=mg$$ when $$\omega = 0$$.

If the planet is made to rotate faster then $$X$$ will decrease

$$mg -R\downarrow =mr\omega^2 \uparrow$$

until the angular speed is such that $$X=0$$, the reading on the scales is zero and the mass is apparently "weightless".

Increasing the speed of rotation of the planet still further will mean that the scales will now be on a ceiling with the mass closer to the planet than the scales and application of Newton's second law gives

$$mg +X =mr\omega^2$$.

If the apparent weight, the reading on the scales, $$X$$, is required to be equal to the true weight $$mg$$ then $$mg+mg=mr\omega^2 \Rightarrow \omega =\sqrt{\dfrac{2g}{r}}$$

• Thanks! This is a very intuitive way of seeing it :) – mar May 24 '19 at 8:24