A sticker is fixed to a huge spherical ball of radius 2m using a gum force of 10N. The spherical ball is rotating at 120 rpm. Find the minimum value of the mass of the spherical ball required so that the sticker doesn't fly off
(Assume the sticker is on the equator line)

- ganeshie8

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- ganeshie8

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- ganeshie8

using newton's second law, \(F_{net}=ma\)
\(10 + \dfrac{GMm}{r^2}=m\omega^2r \)
\(m\) won't cancel, so the question has insufficient info is it ?

- dan815

um

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## More answers

- dan815

im back xD

- dan815

theres enough info

- ganeshie8

im getting two unknowns : \(m,M\)
and i have only one equation to use

- dan815

whats the other M for

- ganeshie8

M : mass of spherical ball
m : mass of sticker

- dan815

oh hmm

- dan815

does that even matter

- ganeshie8

im not really sure how gum force acts..

- dan815

whats the force due to the rotation on a spinning disk again

- ganeshie8

thats same as gravitational force right

- ganeshie8

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- dan815

f=mv^2/r then?

- ganeshie8

\(F_{centripetal}\) is the force causing that centripetal acceleration of the body \(v^2/r\)

- ganeshie8

\(F_{centripetal}\) role is played by the gravitational force in our case

- ganeshie8

oh there is gum force also, which i have never dealt with before..

- dan815

hmmmmmm ya we need mass

- ganeshie8

using newton's second law, \(\sum F =ma\)
\(F_{gum} + F_{centripetal} = ma\)
\(10 + \dfrac{GMm}{r^2}=m\omega^2r \)
\(10 + \dfrac{GMm}{r^2}=m\omega^2r \)

- dan815

but i duno its weird like

- dan815

if u think about lets say big scale

- dan815

if its 2 different people holding a bar

- ganeshie8

im sure that much is correct... but i don't believe there is enough info as we cannot solve 2 unknowns using one equation..

- dan815

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- dan815

like if we think largescale, and these ppl have the same holding force

- dan815

would their pass really play that big of a role

- dan815

mass*

- ganeshie8

i think it should play... thats the principle of centrifuge machine right... used to separate cream from milk...

- dan815

then not enuff info

- dan815

um about this

- dan815

lets suppose there is some arb mass, oky instead of a sticky gum think about an object with mass

- dan815

if it was some object with some mass such that with the gravity of the sphere its 10 Newtons

- dan815

so forget sticky for now

- dan815

then do we have enough info to solve it

- dan815

ok we do butt it doesnt help

- ganeshie8

i think we can find the minimum mass required such that the sticker doesn't flyoff when there is no gum force...

- dan815

u know what silly lol.. the minimum mass can be 0

- ganeshie8

that would be same like the minimum mass that earth should have such that the people on surface of earth don't fly away due to spin of earth

- dan815

then the gum could just attract the sphere and ur done

- ganeshie8

lol lets just say \(M\gt \gt m\) to remove the trivialities :)

- MrNood

if m is taken as 0 then there is no gravitational force to be considered, and hence the answer does not depend on M either......

- MrNood

Also - let's take a reality check here:
we have a force of 10N due to the gum
just taking M = 1000kg and m = 0.1kg as starting points
the resultant gravitational attraction force between the bodies is 1.7 x 10^-9 N
so the gravity force is absolutely negligible in this case
what a weird question - where did it come from?

- ganeshie8

yeah 1000kg is too small to contribute anything... without the gum force, i think the ball must be of size earth or something to keep the sticker from flying off

- MrNood

it says it's 2m radius
even made from the densest material on earth it would be an insignificant forc

- ganeshie8

blackhole or neutron star will do
but it seems the gum force is the thing that keeps the sticker from flying off

- Michele_Laino

here is my reasoning:
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we can write the subsequent vector equation:
\[\Large {{\mathbf{F}}_S} + {{\mathbf{F}}_G} + {{\mathbf{F}}_C} = 0\]
where \(F_S\) is the gum force, \(F_G\) is the gravitational force, and \(F_C\) is the centrifugal force. From such vector equation, we get the subsequent scalar equation:
\[\Large - {F_S} - G\frac{{Mm}}{{{R^2}}} + \frac{{m{\omega ^2}}}{R} = 0\]
where \(R\) is the radius of the large sphere

- ganeshie8

pretty sure it is a typo...
\(\large - {F_S} - G\frac{{Mm}}{{{R^2}}} + \color{red}{m{\omega ^2R}} = 0\)
\(m\) doesn't cancel... so we cannot solve with the given info right ?

- ganeshie8

@MrNood it was asked by my roomate, so there is a high chance for the question to be invalid or having insufficient info...

- MrNood

My hackles start to rise when I see 'centrifugal force'
it is not a force
I accept it comes out with the same equation - but I prefer centripetal force (with opposite sense)
I agree that there is insufficient information
and even if there were - the mass would be unfeasibly large - and related to any sensible 'physics' question - merely science fiction...

- MrNood

^^"unrelated"

- Michele_Laino

yes! you are right! @ganeshie sorry for my typo!

- ganeshie8

I think michele is setting up that equation in rotational frame of reference, so centrifugal force, even if it is a pseudo force, exists right ?

- Michele_Laino

correct! @ganeshie8

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