Consider a mass moving with velocity v0 radius r0 on a table top by holding a string attached to the mass and passing through the centre of the table into my hand below the table.
Suppose I decrease radius to r then what is the value of final velocity v?
1)Is it given using Thand=mv2/r=mv02r0 or using
2)torque ext =0 =>angular momentum conserved =>mv0r0=mvr or using
3)work-energy theorem => Thand(r0-r)=1/2m(v2-v02)
Plz answer my doubt.
well the first eqn is not right since the tension in the hand will definitely change.
the second one is correct
for the third one since the tension is not constant and for the integration also there will be two unknowns therefore the third eqn ids not correct hence onle angular moimentum can be coserved as it causes no torque
P.S :did u get cleared on the ionic douubt that u had if u hav then pleese explain it to me too
I can keep tension const and reduce r as then v increases.
I don't understand why you say that T changes
Well I wrote wrote wrong in my previous reply.
I wanted to ask that T can remain const while r & v variate
that is
u want to say that if r decreases then v also decreases but that can not be true as there is no external torque henc mvr=constant i.e is r decreases then v should increase
dear ankul
acc. to the first eqn vincreases as r increases but actually it should decrease .Also in any of the cases there isnt any net torque hence shouldny anguar mom. be conserved?
Let me assume that initial velocity & radius are Vi and Ri. & the final velcity and radius are Vf and Rf .
Since u pull the string and hence the force and radius vector are along the same line therefore no net torque acts on the system and hence angular momentum is conserved. Thus equation 2 must be used.
It is possible to pull the string in such a way that the initial and final tensions remain the same. In such a case both equations 1) and 2) are to be used and hence the values of Rf and Vf are fixed (meaning: u cannot choose the Rf and the fact that tension remains the same. Choosing one automatically determines the other.)
Out of the manny possible ways in which the radius can be reduced one of them is the way in which the tension of the string remains the same.
Ur equation 3) is completely wrong and hence cannot be used. The reason is that the radius cannot be reduced without a spiral motion. In this case we seem to simplify the problem and take into account the initial and final states only. When u apply the extra tension to the body directed to the centre u set it in motion along the radius. Now by conservation of linear momentum along the radius how can u explain its abrupt stopping at the desired radial distance. Obviously a retarding force must act in the direction of the radius to bring the body to a stop. This complicates the situation and hence work energy theorem cannot be applied with the simplification u seek.
Hey Asif,
I understood your point.But then my point is valid in case if i quasi-satically reduce radius from Ri to Rf so that Thand almost remains const. through-out.
then work energy just says all kinds of work(int.+ext.+non-conservative)=delta(K).Here all kinds of work=Thand x(Rf-Ri ) in which case even 3rd eq is valid.
Hey,
Can you please explain what quasi-statically means.