SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 30

30. Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_{1}$ has $2 k g$ mass and $0.2 m$ radius and initial angular velocity of $50 r a d s^{- 1}$ . Disc $D_{2}$ has $4 k g$ mass, $0.1 m$ radius and initial angular velocity of $200 r a d s^{- 1}$ . The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in $r a d s^{- 1}$ ) of the system is [NEET Kar. 2013]

(a) 40

(b) 60

(d) 120

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Answer:

Correct Answer: 30. (c)

Solution:

$m_{2} = 4 k g$

$r_{1} = 0.2 m$

$r_{2} = 0.1 m$

$ω_{1} = 50 r a d s^{- 1}$

$ω_{2} = 200 r a d s^{- 1}$

As, angular momentum,

$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{1}$

$∴ ω_{f} = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}} = \frac{\frac{1}{2} m_{1} r_{1}^{2} ω_{1} + \frac{1}{2} m_{2} r_{2}^{2} ω_{2}}{\frac{1}{2} m_{1} r_{1}^{2} + \frac{1}{2} m_{2} r_{2}^{2}}$

By putting the value of $m_{1}, m_{2}, r_{1}, r_{2}$ and solving we get $= 100 r a d s^{- 1}$

System of Particles and Rotational Motion - Result Question 30

Answer:

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System of Particles and Rotational Motion - Result Question 30

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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