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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 May 2007 15:23:23 IST
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a wheel is rolling whith constant speed. what is the diarection of acceleration of point of contact with earth?
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21 May 2007 15:30:01 IST
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since wheel is rolling with constant speed so, i think acceleration of any point/particle will be towards center i.e. radial acceleration...
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21 May 2007 17:16:19 IST
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I agree with arnie. As the velocity of rolling is constant, its acceleration is towards centre.
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21 May 2007 19:37:35 IST
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Point of contact of a rolling wheel is has got its speed = 0 So the tangential acceleration is always = 0 but there is a centripetal acceleration = v2 /R where v = speed of center of point R = radius of the wheel.
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21 May 2007 23:18:43 IST
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as the wheel is rolling without slipping i feel that the acceleration of the lowest point is at every instance at rest wrt the earth hence its acceleration is =0
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21 May 2007 23:31:11 IST
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Its speed is zero and not the acceleration...!!
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22 May 2007 04:13:41 IST
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u r right neeraj...................the tangentil acc is only zero.it has radial component...................................................
v==r (cross) w
differentiate
then u get that the acc has two components.
these comp are perpendicular to each other.
tangential cop =0.
so the effetive accc = radial copmanent
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22 May 2007 06:46:08 IST
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Its speed is zero and not the acceleration...!!
How the speed will be zero?It's linear acc should be zero isn't it?Plz clear this doubt.
a wheel is rolling whith constant speed
Vinodh
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22 May 2007 13:46:17 IST
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you must have read about it...
when we throw a ball up in the air ...at the topmost point its speed is 0...but it has some downward acceleration(g)...
In this case the lowermost point is at rest......but its acceleration is towards the centre...(centripetal)
it is moving with constant speed means its TANGENTIAL acceleration is 0
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22 May 2007 13:49:44 IST
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OHhh...no..i think i m wrong...
The acc. vector has two components...one from the linear acceleration and the other is w^2. r.
they cancel each other if the wheel is rolling without slipping...
The acceleration is towards the centre if the wheel is not moving linearly but its is moving on a curved path...
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22 May 2007 15:14:20 IST
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what will be the accelaration of the point of contact 'O' of the wheel, rolling without slipping, aftrer time='t's if the accelaration of wheel is constant & is equal to= 'a'm/s2? plz give a general solution for this.
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22 May 2007 15:18:24 IST
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@kritika i think it shd be a vector sum of --- a nd centripital acceleration
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22 May 2007 15:39:23 IST
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Mr.coolyog, for the point of contact 'O' of d wheel,linear acc.=0 so its acc shall be= a2t2/r,(centripetal acc calculated using given qty.s) where r is radius, a=acc of wheel, t=time ta which acc of 'o' is calculated. what do u say?
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23 May 2007 21:39:05 IST
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The acc. vector has two components...one from the linear acceleration and the other is w^2. r.
they cancel each other if the wheel is rolling without slipping...
The acceleration is towards the centre if the wheel is not moving linearly but its is moving on a curved path...
Can u plz xplain this with a Simple figure?
Vinodh
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24 May 2007 17:19:35 IST
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See ...
a wheel is rotating ...towards right.
It has some linear velocity(v) and some angular velocity(w).
At lower most point, one component of velocity is towards right(v)...and one component (rw) is towards left.
If the components are equal ,the lower most point is at rest.
similarly...acc is 0
If the components are not equal...the wheel is rolling with slipping...
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