a toy car tied to the end of an unstretched string of length l when revolved, the toy car moves in horizontal circle with radius 2l and time period T. if it's speeded until it moves in horizontal circle of radius 3l with period T1, Tejgaon between T and T1 is (Hooke's law is obeyed)
Answers
The new time period is given by T₁ = √3T/2
Explanation:
Case I
string length = l
string is stretched to 2l, so net increase in length = 2l - l = l
=> spring force, F = kl
where k = spring constant
when the string is rotated with 2l radius,
we know that centripetal force is given by,
F = mrω² = m x 2l x (2π/T)²
Since the centripetal force and spring force should be equal, hence,
kl = m x 2l x (2π/T)²
=> k = 2m x (2π/T)².............eq1
Case II
string length = l
string is stretched to 3l, so net increase in length = 3l - l = 2l
=> spring force, F = k x 2l = 2kl
when the string is rotated with 3l radius,
we know that centripetal force is given by,
F = mrω₁² = m x 3l x (2π/T₁)²
Since the centripetal force and spring force should be equal, hence,
2kl = m x 3l x (2π/T₁)²
=> k = 3m/2 x (2π/T₁)².............eq2
equating eq1 and eq2 we get:
2m x (2π/T)² = 3m/2 x (2π/T₁)²
=> T₁²/T² = 3/4
=> T₁ = √3T/2
Hence the new time period is given by T₁ = √3T/2
The answer is provided in the attachment.
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