
Arturo O. answered 12/07/16
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I assume you mean without changing the tangential speed v.
T1 = mv2/r1
At same tangential speed v but at r1/2,
T2 = mv2/(r1/2) = 2mv2/r1 = 2T1
The tension is doubled if the radius is halved at constant tangential speed. However, such a problem is not very realistic, because if you pull in the string to make the radius smaller, angular momentum is conserved, and the tangential speed will actually increase. Conservation of angular momentum will make
mv1r1 = mv2r2 ⇒ v1r1 = v2r2
If r2 = r1/2, then
v1r1 = v2(r1/2) ⇒ v2 = 2v1
Then
T2 = mv22/r2 = m(2v1)2/(r1/2) = 8mv12/r1 = 8T1
So realistically, the inward force should increase by a factor of 8 if you reduce the radius by half. But in this problem, as stated, with constant tangential speed v, the inward force is doubled.