ganeshie8
  • ganeshie8
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)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
|dw:1447490751122:dw|
ganeshie8
  • 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
  • dan815
um

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dan815
  • dan815
im back xD
dan815
  • dan815
theres enough info
ganeshie8
  • ganeshie8
im getting two unknowns : \(m,M\) and i have only one equation to use
dan815
  • dan815
whats the other M for
ganeshie8
  • ganeshie8
M : mass of spherical ball m : mass of sticker
dan815
  • dan815
oh hmm
dan815
  • dan815
does that even matter
ganeshie8
  • ganeshie8
im not really sure how gum force acts..
dan815
  • dan815
whats the force due to the rotation on a spinning disk again
ganeshie8
  • ganeshie8
thats same as gravitational force right
ganeshie8
  • ganeshie8
|dw:1447492323664:dw|
dan815
  • dan815
f=mv^2/r then?
ganeshie8
  • ganeshie8
\(F_{centripetal}\) is the force causing that centripetal acceleration of the body \(v^2/r\)
ganeshie8
  • ganeshie8
\(F_{centripetal}\) role is played by the gravitational force in our case
ganeshie8
  • ganeshie8
oh there is gum force also, which i have never dealt with before..
dan815
  • dan815
hmmmmmm ya we need mass
ganeshie8
  • 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
  • dan815
but i duno its weird like
dan815
  • dan815
if u think about lets say big scale
dan815
  • dan815
if its 2 different people holding a bar
ganeshie8
  • 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
  • dan815
|dw:1447492672141:dw|
dan815
  • dan815
like if we think largescale, and these ppl have the same holding force
dan815
  • dan815
would their pass really play that big of a role
dan815
  • dan815
mass*
ganeshie8
  • ganeshie8
i think it should play... thats the principle of centrifuge machine right... used to separate cream from milk...
dan815
  • dan815
then not enuff info
dan815
  • dan815
um about this
dan815
  • dan815
lets suppose there is some arb mass, oky instead of a sticky gum think about an object with mass
dan815
  • dan815
if it was some object with some mass such that with the gravity of the sphere its 10 Newtons
dan815
  • dan815
so forget sticky for now
dan815
  • dan815
then do we have enough info to solve it
dan815
  • dan815
ok we do butt it doesnt help
ganeshie8
  • ganeshie8
i think we can find the minimum mass required such that the sticker doesn't flyoff when there is no gum force...
dan815
  • dan815
u know what silly lol.. the minimum mass can be 0
ganeshie8
  • 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
  • dan815
then the gum could just attract the sphere and ur done
ganeshie8
  • ganeshie8
lol lets just say \(M\gt \gt m\) to remove the trivialities :)
MrNood
  • 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
  • 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
  • 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
  • MrNood
it says it's 2m radius even made from the densest material on earth it would be an insignificant forc
ganeshie8
  • 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
  • Michele_Laino
here is my reasoning: |dw:1447498635342:dw| 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
  • 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
  • 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
  • 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
  • MrNood
^^"unrelated"
Michele_Laino
  • Michele_Laino
yes! you are right! @ganeshie sorry for my typo!
ganeshie8
  • 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
  • Michele_Laino
correct! @ganeshie8

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