Question

2 stones of masses m and 2m are whirled in horizontal circles the heavier one in a radius r/2 and the lighter one in radius r the tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. the value of n is
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Answer 1

Answer:

Let v be tangential speed of heavier stone.  

Then, centripetal force experienced by lighter stone  

is (Fc)lighter=m(nv)^2r  

and that of heavier stone is  

(Fc)heavier=2mv^2(r/2)  

But (Fc)lighter=(Fc)heavier (given)  

∴ m(nv)^2r=2mv^2(r/2)  

n^2(mv^2r)=4(mv2r)  

n^2=4 or n = 2

Answer 2

Answer:

2 is the value of n.

Explanation:

For the heavier stone, If we take that v is the tangential speed.

So, the centripetal force which tye lighter stone experienced will be

Lighter stone F=m*(nv)^2r.

While for that of the heavier stone will be

F for heavier=2m*v^2(r/2)  

Since, from the question we get that force for lighter=force for heavier stone.

Therefore, m*(nv)^2r=2*m*v^2(r/2).

So, n^2(m*v^2r)=4(m*v^2r).

Hence, on solving we will get the value of

n^2=4 or n = 2.