Angular momentum conservation
I1ω1+I2ω2=(I1+I2)ω
ω=I1+I2I1ω1+I2ω2
Loss =21I1ω12+21I2ω22−21(I1+I2)ω2
=21I1ω12+21I2ω22−21(I1+I2)(I1+I2I1ω1+I2ω2)2
=21(I1+I2)I1I2(ω1−ω2)2
Ei−Ef=2(I1+I2)I1I2(ω1−ω2)2
Two discs have moments of intertia 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:
Held on 27 Aug 2021 · Verified 6 Jul 2026.
(I1+I2)I1I2(ω1−ω2)2
2(I1+I2)(ω1−ω2)2
2(I1+I2)I1I2(ω1−ω2)2
2(I1+I2)(I1−I2)2ω1ω2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.