Rotational Motion Question 28
Question: A stone of mass m, tied to the end of a string, is whirled around in a circle on a horizontal frictionless table. The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by $ T=Ar^{n}, $ where A is a constant, r is the instantaneous radius of the circle. The value of n is equal to
[JEE ONLINE 26-05-2012]
Options:
A) - 1
B) - 2
C) - 4
D) - 3
Show Answer
Answer:
Correct Answer: D
Solution:
Angular momentum is constant
⇒$mr^2 ω=const⇒ ω= \frac{const}{mr^2} $
$T=mω^2 r=m({\frac{const}{mr^2}})^2 $
$r=( const )r^{−3} $
thus, n=−3
BETA
