A disc of radius R and mass M is rolling horizontally without slipping with speed v . It then moves

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6.System of Particles and Rotational Motion

hard

A disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

$\frac{v^2}{g}$

$\frac{3}{4} \frac{v^2}{g}$

$\frac{1}{2} \frac{v^2}{g}$

$\frac{2}{3} \frac{v^2}{g}$

(JEE MAIN-2024)

Solution

Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational $\mathrm{KE}$ remains unchanged as there is no friction.

$\frac{1}{2} \mathrm{mv}^2=\mathrm{mgh}$

$\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}$

Standard 11

Physics

Two bodies have their moments of inertia $I$ and $2 I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio

medium

(AIPMT-2005)

Two point masses of $0.3\ kg$ and $0.7\ kg$ are fixed at the ends of a rod of length $1.4\ m$ and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of

medium

(AIIMS-2000)

A ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is

easy

(AIPMT-1988)

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent)

normal

A circular disc is rolling on a horizontal plane. Its total kinetic energy is $300\,\,J.$ ……… $J$ is its translational $K.E.$