ParthKohli
  • ParthKohli
Find the minimum height of the obstacle so that the sphere remains in equilibrium.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Astrophysics
  • Astrophysics
.
ParthKohli
  • ParthKohli
|dw:1447046974800:dw|
ganeshie8
  • ganeshie8
obstacle is fixed to the ramp is it ?

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More answers

ParthKohli
  • ParthKohli
yes
ganeshie8
  • ganeshie8
Okay, then I think we want the torque about the point \(P\) to be in counter clockwise direction for equilibrium : |dw:1447047180182:dw|
ganeshie8
  • ganeshie8
then that counterclockwise torque would be balanced by the normal force from the ramp
ganeshie8
  • ganeshie8
so i think we simply need to see when the torque becomes 0
ganeshie8
  • ganeshie8
having dealt with the physics part, rest is just geometry... let me see if i can set up the equations correctly..
lochana
  • lochana
is the surface rough?
ganeshie8
  • ganeshie8
friction shouldn't matter for this problem..
lochana
  • lochana
okay:)
ganeshie8
  • ganeshie8
we need to check when that gravity force aligns over the pivot point \(P\)
ParthKohli
  • ParthKohli
you did it, haha
ganeshie8
  • ganeshie8
|dw:1447049974188:dw|
ganeshie8
  • ganeshie8
|dw:1447050753241:dw|
ganeshie8
  • ganeshie8
@lochana friction dosn't matter here because, the sphere is not sliding or accelerating along the surface of the ramp. for breaking equilibrimum, we are lifting the sphere about the point P. while doing that, the sphere gets rotated "perpendicular" to the ramp, about the point P. notice that the sphere is not accelerating along the surface of ramp... so there won't be any frictional force on the sphere... thats my understanding.. i could be wrong though...
ganeshie8
  • ganeshie8
i think below statement is true, but its not so obvious w/o messing with the math : ``` friction is 0 in rolling motion when the velocity is constant friction acts only when the velocity changes ```
ParthKohli
  • ParthKohli
yes, friction is zero because it acts only when relative motion occurs between the bottommost point and the surface. in pure rolling no friction acts if we leave a sphere undisturbed.
ganeshie8
  • ganeshie8
Ahh right, when the sphere is in pure rolling motion, the velocity of bottommost point, relative to the ramp is 0
ganeshie8
  • ganeshie8
how do you explain why the friction exists when the sphere accelerates ?
ParthKohli
  • ParthKohli
|dw:1447052626746:dw|
ParthKohli
  • ParthKohli
When \(v_{CM}\) increases or decreases (acceleration), friction acts to oppose relative motion between the bottommost point. It acts in a way that \(R\omega\) becomes equal to \(v_{CM}\) again.
ganeshie8
  • ganeshie8
|dw:1447052924655:dw|
ganeshie8
  • ganeshie8
is that right ?
ParthKohli
  • ParthKohli
that direction is seen when the wheel is rotating way too fast and is translating slowly. so static friction acts forward in order that the translational velocity increases, rotational velocity decreases, and rolling is attained again.
ganeshie8
  • ganeshie8
what would be the direction of static friction in our case, if the obstacle is removed and sphere starts to accelerate along the ramp under the influence of gravity ?
ganeshie8
  • ganeshie8
would it be forward or backward ?
ParthKohli
  • ParthKohli
backwards
ganeshie8
  • ganeshie8
why
ParthKohli
  • ParthKohli
because see, it tends to translate downwards due to gravity, but gravity does not provide any torque to the wheel so as to make it rotate. so it needs something to rotate clockwise, which is provided by the static friction of course.
ganeshie8
  • ganeshie8
gravity produces no torque about the center of mass of sphere
ParthKohli
  • ParthKohli
exactly, so friction has to do that job
ganeshie8
  • ganeshie8
isn't it conflicting with the previous explanation of why frction must be in the same direction as acceleration of the CM ? |dw:1447053505383:dw|
ParthKohli
  • ParthKohli
i never said that static friction is in the same direction as v_cm.
ganeshie8
  • ganeshie8
on a horizontal ramp, if the sphere is accelerating, the static friction is in the same direction is v_cm right ?
ParthKohli
  • ParthKohli
not at all... it depends. think about it this way: if a wheel is being rotated but not translated by any force, then it needs friction to translate it. if a wheel is being translated but not rotated by any force, then it needs friction to rotate it. in absence of friction, this is the second case, right? because gravity is translating it and no other force is available to rotate it other than friction.
ParthKohli
  • ParthKohli
|dw:1447053732852:dw|
ParthKohli
  • ParthKohli
this is the first case ^ think about the direction of friction here
ganeshie8
  • ganeshie8
that is true only when "forces" act on the wheel friction will be 0 when the center-of-mass of the wheel moves with constant velocity
ganeshie8
  • ganeshie8
|dw:1447054878534:dw|
lochana
  • lochana
So how we get inequalities for the minimum height? what is the argue?
ganeshie8
  • ganeshie8
|dw:1447055165940:dw|
lochana
  • lochana
@ganeshie8 agreed!
ganeshie8
  • ganeshie8
|dw:1447055466285:dw|
lochana
  • lochana
are sure about gravity and reaction at p on the same line and opposite?
ganeshie8
  • ganeshie8
are you asking why must the force of gravity pass through the point P for minimum height ?
ganeshie8
  • ganeshie8
|dw:1447055700029:dw|
lochana
  • lochana
yes. ah okay. It is possible though. After that that equilibrium, it is definitely going to rotate.
ganeshie8
  • ganeshie8
exactly! check this http://physics.stackexchange.com/questions/258/what-determines-the-minimum-angle-at-which-a-domino-falls-over
ganeshie8
  • ganeshie8
https://youtu.be/yNQi5g8JfPg A bob is suspended from a string connected to the block at it's center of mass. Raise the inclined plane until the block tips over. Observe that the block does not tip over until the center of mass exceeds the base of the block, illustrated when the string and mass clear the bottom edge of the block.

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