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Rotational Motion question

2020 · 9 Jan · Shift 2 · Q50
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Rotational Motion question

2020 · 9 Jan · Shift 2 · Q50

JEE MainPhysicsRotational MotionMCQ+4 / −1
A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses m1 and m2 (m1 >\gt> m2) are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when m1 descents by a distance h is : JEE Main 2020 (Online) 9th January Evening Slot Physics - Rotational Motion Question 149 English
  1. A
    [2(m1+m2)gh(m1+m2)R2+I]12{\left[ {{{2\left( {{m_1} + {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}[(m1​+m2​)R2+I2(m1​+m2​)gh​]21​
  2. B
    [m1+m2(m1+m2)R2+I]12gh{\left[ {{{{m_1} + {m_2}} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh[(m1​+m2​)R2+Im1​+m2​​]21​gh
  3. C
    [(m1−m2)(m1+m2)R2+I]12gh{\left[ {{{\left( {{m_1} - {m_2}} \right)} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh[(m1​+m2​)R2+I(m1​−m2​)​]21​gh
  4. D
    [2(m1−m2)gh(m1+m2)R2+I]12{\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}[(m1​+m2​)R2+I2(m1​−m2​)gh​]21​
View written solutionFree

Correct answer: D

  1. Set up the energy change

When the system is released, m1m_1m1​ moves downward by hhh and m2m_2m2​ moves upward by hhh.

So, the net decrease in gravitational potential energy is

ΔU=m1gh−m2gh=(m1−m2)gh\Delta U = m_1gh - m_2gh = (m_1-m_2)ghΔU=m1​gh−m2​gh=(m1​−m2​)gh

Since the string does not slip on the wheel,

v=Rωv = R\omegav=Rω

where vvv is the linear speed of the blocks and ω\omegaω is the angular speed of the wheel.

  1. Write total kinetic energy gained

Both blocks have speed vvv, so their translational kinetic energy is

Kblocks=12m1v2+12m2v2=12(m1+m2)v2K_{\text{blocks}} = \frac12 m_1v^2 + \frac12 m_2v^2 = \frac12 (m_1+m_2)v^2Kblocks​=21​m1​v2+21​m2​v2=21​(m1​+m2​)v2

The wheel has rotational kinetic energy

Kwheel=12Iω2K_{\text{wheel}} = \frac12 I\omega^2Kwheel​=21​Iω2

Hence total kinetic energy is

K=12(m1+m2)v2+12Iω2K = \frac12 (m_1+m_2)v^2 + \frac12 I\omega^2K=21​(m1​+m2​)v2+21​Iω2

Using v=Rωv=R\omegav=Rω,

K=12(m1+m2)R2ω2+12Iω2K = \frac12 (m_1+m_2)R^2\omega^2 + \frac12 I\omega^2K=21​(m1​+m2​)R2ω2+21​Iω2 K=12[(m1+m2)R2+I]ω2K = \frac12 \left[(m_1+m_2)R^2 + I\right]\omega^2K=21​[(m1​+m2​)R2+I]ω2
  1. Apply conservation of mechanical energy

The loss in potential energy equals the gain in kinetic energy:

(m1−m2)gh=12[(m1+m2)R2+I]ω2(m_1-m_2)gh = \frac12 \left[(m_1+m_2)R^2 + I\right]\omega^2(m1​−m2​)gh=21​[(m1​+m2​)R2+I]ω2

So,

ω2=2(m1−m2)gh(m1+m2)R2+I\omega^2 = \frac{2(m_1-m_2)gh}{(m_1+m_2)R^2 + I}ω2=(m1​+m2​)R2+I2(m1​−m2​)gh​

Therefore,

ω=[2(m1−m2)gh(m1+m2)R2+I]1/2\boxed{\omega = \left[\frac{2(m_1-m_2)gh}{(m_1+m_2)R^2 + I}\right]^{1/2}}ω=[(m1​+m2​)R2+I2(m1​−m2​)gh​]1/2​
  1. Compare with options

This matches Option D:

[2(m1−m2)gh(m1+m2)R2+I]1/2{\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{1/2}}[(m1​+m2​)R2+I2(m1​−m2​)gh​]1/2

So the correct answer is:

D\boxed{\text{D}}D​
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