KTUYSBeta
Log inSign up

System of Particles and Rotational Motion · Class 11 · CBSE · Science Physics

A ballet dancer spins with angular velocity omega0 on a frictionless turntable with her arms outstretched. If she folds her arms such that her moment…

Practice question from System of Particles and Rotational Motion (Class 11 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

easy · MCQ · Q-79503A

A ballet dancer spins with angular velocity ω0\omega_0 on a frictionless turntable with her arms outstretched. If she folds her arms such that her moment of inertia decreases to 75 %75\text{ \%} of its initial value, her new angular velocity becomes:

  1. A.0.75ω00.75 \omega_0
  2. B.1.33ω01.33 \omega_0Correct answer
  3. C.1.50ω01.50 \omega_0
  4. D.2.00ω02.00 \omega_0

Solution

Due to conservation of angular momentum, I1ω1=I2ω2I_1 \omega_1 = I_2 \omega_2. Given I2=0.75I1=34I1I_2 = 0.75 I_1 = \frac{3}{4} I_1, we have I1ω0=(34I1)ω2  ⟹  ω2=43ω0≈1.33ω0I_1 \omega_0 = \left(\frac{3}{4} I_1\right) \omega_2 \implies \omega_2 = \frac{4}{3} \omega_0 \approx 1.33 \omega_0.

More from System of Particles and Rotational Motion

All of System of Particles and Rotational Motion

Practise more Physics

Sign up free for adaptive practice on this chapter, with instant feedback and hints on every question.

Start practising free

Free to start · No credit card needed