Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1 and ω2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

Using conservation of angular momentum we getI1ω1+I2ω2=I1+I2ω⇒ω=I1ω1+I2ω2I1+I2Initial Kinetic energyKi=12I1ω12+12I2ω22Final K.E kf=12I1+I2ω2Loss in Kinetic energy=Ki−Kf=12I1I2I1+I2ω1−ω22

Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1 and ω2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

Using conservation of angular momentum we getI1ω1+I2ω2=I1+I2ω⇒ω=I1ω1+I2ω2I1+I2Initial Kinetic energyKi=12I1ω12+12I2ω22Final K.E kf=12I1+I2ω2Loss in Kinetic energy=Ki−Kf=12I1I2I1+I2ω1−ω22

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two discs have moments of inertia I₁ and I₂ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $ω$ ₁ and $ω$ ₂ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{I_{1} I_{2}}{(I_{1} + I_{2})} {(ω_{1} - ω_{2})}^{2}$

$\frac{{(I_{1} - I_{2})}^{2} ω_{1} ω_{2}}{2 (I_{1} + I_{2})}$

$\frac{{(ω_{1} - ω_{2})}^{2}}{2 (I_{1} + I_{2})}$

$\frac{I_{1} I_{2}}{2 (I_{1} + I_{2})} {(ω_{1} - ω_{2})}^{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Using conservation of angular momentum we get
$\begin{array}{l} I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω \\ \Rightarrow ω = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}} \end{array}$
Initial Kinetic energy
$K_{i} = \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} I_{2} ω_{2}^{2}$
Final K.E $k_{f} = \frac{1}{2} (I_{1} + I_{2}) ω^{2}$
Loss in Kinetic energy
$= K_{i} - K_{f} = \frac{1}{2} (\frac{I_{1} I_{2}}{I_{1} + I_{2}}) {(ω_{1} - ω_{2})}^{2}$