# DOES ANY TORQUE ACTS ON A FREELY FALLING BODY about axis some distance away from its point of fall on earth???

A particle is falling freely due to gravity.

what is the torque acting on it due to gravitational force about an axis 1 meter left from its point of freefall on earth???plz calculate

If torque acts on particle then WHY any angular acceleration is not produced in it.(OR DOES IT PRODUCED)???

Update:

problem is that u find a constant value of net torque acting on the particle(plz verify that urself).but it is not seen to be 'rotating' from the earth.

So why is it not rotating despite a TORQUE ACTING ON IT.

Am I not able to locate the 'torque produced' or I am calculating the torque incorrectly (or that probably somehow its wrong to calculate torque like this)

and 'Hector P' considering the motion of stone with earth atmosphere it is always moving in a same direction.bot torque on same particle comes to be different if we take different axes.plz explain how U are considering it as the LAG EFFECT.

and 'stronger_..' if gravity is a weak force still then also is there any effect of this torque however small.

Update 2:

Mr. steve,ur answer seems to be the one I am looking for.but what exactly is 'd' in ur equation "T = F*d".distance between the Centre of mass and the centre of gravity or distance between the Centre of gravity and reference point on earth.plz clarify,i'll be greatful.

and kaski guy u hav a point.but i always try to keep it as simple as possible.that's why I put phrase "freely falling body" in main body of question and then I used 'particle' in ad. details to explain what my question exactly is.but i'll keep ur advice in mind in future questions.I beg ur pardon.

and "pearlsawmme" U said that "The center of gravity and center of mass coincide for all objects.Hence there is no resultant torque acting on the freely falling body" U did not give any reason that why this is so I mean that why (or how) coincidence of the C.M. and C.G. of a body makes net torque zero on it.

and 'stronger_..' thanx fr that excellant link but I asked how we qualitatively explain the effect,be it however small

Relevance
• Anonymous

the torque on the particle is force X perpendicular distance of axis from line of fall of particle.

so it is mass X g X 1m.

now as we find a torque about axis then must have some angular acceleration.and magnitude of this angular acceleration is

torque on particle / moment of inertia about that point (mass X square of 'distance' between reference point and the particle).

'distance' between reference point and the particle is the actual separation between particle and reference point of axis, NOT the perpendicular distance which we took before.

this distance is variable so the angular acceleration must be a variable also.how do we can find this variable acceleration.

resolve the acceleration due to gravity in 2 components.one along the line joining the reference point and particle and other perpendicular to it.

this perpendicular component is what we see due to torque.

problem here is that we forget the role of 'moment of inertia' and struck only by the 'constant' value of torque.

and some answers said that 'since the CM and CG always coincide the torque about gravity is 0'

but this statement is true only if we consider the torque about the CM of particle.not when we take axis as given in this problem.

this may be hard to visualize a bit but try it in copy with the proper diagrams and U will understand that working.

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• For any object we can find the center of mass.

The center of mass is the point where all masses can be thought of to be concentrated.

When we consider the gravitational force acting on an

object, if we find the resultant of all forces, the resultant is acting on a point called center of gravity.

The center of gravity and center of mass coincide for all objects.

Hence there is no resultant torque acting on the freely falling body.

If we have given some angular momentum initially for the freely falling body, then as per conservation of angular momentum the body will maintain its angular momentum. Again there is no resultant torque acting on this.

For a freely falling body if it had initial angular momentum, the body will rotate about any of the two free axes, which every object is having when allowed to rotate freely.

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• There is no torque acting on the particle (or body). The gravitational force (the ONLY one present) acts through the center of gravity and therefore in the formula T = F*d, d = 0 so there IS NO TORQUE.

Most of us understand this intuitively.

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• the torque formula

T=r X f

where f=weight (mg) which acts down and r is the vector

position of the particle as it falls, relative to the axis (for

simplicity let's assume that the axis is perpendicular to

the plane containing the motion and the origin) gives

T=mg|r|sin(theta)

where theta is the angle between r and f.

so, because

|r| sin(theta)= 1 meter,

we obtain

T=mg times 1 meter

which is constant

note that the torque depends on the perpendicular

distance between the axis chosen and the line along

which the particle is falling. the torque is only zero if the

particle falls along a line that passes through the axis.

i believe the angular acceleration you are looking for is

the second time derivative of theta (basically the angle

swept out by the line from the origin to the particle)

think of it as free fall expressed in polar coordinates.

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• Anonymous

The earth would be rotating and although the atmosphere will carry it along, I believe there would be a lag effect, much like when you throw something out of a car window.

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• Anonymous

I’ve read your questions. What an inquisitive mind! And I mean it.

What do you mean by particle? Why not to say just: a boulder is falling freely from a cliff? The question would look less vague this way! clarify.

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• gravity is a very weak universal force.

it's effects on that small a scale are negligible .

funny I was looking for the effects of bending space and time and came across a website that had a calculator that could help

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