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Rotational Motion question

2015 · Q144
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Rotational Motion question

2015 · Q144

NEETPhysicsRotational MotionMCQ+4 / −1
A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. the tension in the string is increased gradually and finally m moves in a circle of radius R02{{{R_0}} \over 2}2R0​​.

The final value of the kinetic energy is
AIPMT 2015 Cancelled Paper Physics - Rotational Motion Question 74 English
  1. A
    2mv02_0^202​
  2. B
    12{1 \over 2}21​mv02_0^202​
  3. C
    mv02_0^202​
  4. D
    14{1 \over 4}41​mv02_0^202​
View written solutionFree

Correct answer: A

According to law of conservation of angular momentum

mvr = mv'r'

v0R0=v(R02);v=2v0{v_0}{R_0} = v\left( {{{{R_0}} \over 2}} \right);v = 2{v_0}v0​R0​=v(2R0​​);v=2v0​   …(i)

∴\therefore∴ K0K=12mv0212mv2=(v0v)2{{{K_0}} \over K} = {{{1 \over 2}mv_0^2} \over {{1 \over 2}m{v^2}}} = {\left( {{{{v_0}} \over v}} \right)^2}KK0​​=21​mv221​mv02​​=(vv0​​)2

⇒\Rightarrow⇒ KK0=(vv0)2=(2)2{K \over {{K_0}}} = {\left( {{v \over {{v_0}}}} \right)^2} = {\left( 2 \right)^2}K0​K​=(v0​v​)2=(2)2    (Using (i))

K = 4K0 = 2mv02{mv_0^2}mv02​

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