A ring of mass m is rolling without slipping& | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{KE}=\frac{1}{2} \mathrm{mV}_{\mathrm{C}}^{2}+\frac{1}{2} \mathrm{I}_{\mathrm{C}} \omega^{2}$

$=\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \ma

Vedclass · Accepted Answer

$\frac{5}{3}m{v^2}$

Vedclass · Answer

$\mathrm{KE}=\frac{1}{2} \mathrm{mV}_{\mathrm{C}}^{2}+\frac{1}{2} \mathrm{I}_{\mathrm{C}} \omega^{2}$

$=\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \mathrm{mR}^{2}\left(\frac{\mathrm{V}^{2}}{\mathrm{R}^{2}}\right)+\frac{1}{2} \mathrm{mV}^{2}+$

$\frac{1}{2} \frac{\mathrm{m}\left(2 \mathrm{R}^{2}\right)}{12} \frac{\mathrm{V}^{2}}{\mathrm{R}^{2}}$

$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown is figure. $A$ rod of identical mass is fixed along one of its diameter. The total kinetic energy of the system is :-

Similar Questions

Ratio of total energy and rotational kinetic energy in the motion of a disc is

Which of the following (if mass and radius are assumed to be same) have maximum percentage of total $K.E.$ in rotational form while pure rolling?

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \,\%$ of its translational kinetic energy. The body is :

A smooth tube of certain mass closed at both ends is rotated in a gravity free space and released. The two balls shown in figure moves towards the ends of the tube and stay there. Then which statement is incorrect about this whole system