Which of the following (if mass and radius ar | vedclass

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

Question

(c)

$K _{	ext {Rother }}= K _{	ext {Trenstation }}+ K _{	ext {rotaton }}$

$=\frac{1}{2} m v^2+\frac{1}{2} l \omega^2$

$=\frac{1}{2} m R^2

Vedclass · Accepted Answer

Ring

Vedclass · Answer

(c)

$K _{	ext {Rother }}= K _{	ext {Trenstation }}+ K _{	ext {rotaton }}$

$=\frac{1}{2} m v^2+\frac{1}{2} l \omega^2$

$=\frac{1}{2} m R^2 \omega^2+\frac{1}{2}\left|\omega^2=\frac{1}{2}ight| \omega^2\left(1+\frac{m R^2}{1}ight)$

$\Rightarrow$ ratio $=\frac{K_{	ext {waton }}}{K_{	ext {total }}}=\frac{\frac{1}{2} I ^2}{\frac{1}{2} I_{\omega^2}\left(1+\frac{m R^2}{1}ight)}$

$=\frac{1}{1+\frac{m R^2}{1}}$

For disc,

$	ext { Ratio }=\frac{1}{1+\frac{m R^2}{\left(\frac{m R^2}{2}ight)}}=\frac{1}{3}$

Solid sphere

$	ext { Ratio }=\frac{1}{1+\frac{m R^2}{\frac{2}{5} m R^2}}=\frac{2}{7}$

Ring

$	ext { Ratio }=\frac{1}{1+\frac{m R^2}{m R^2}}=\frac{1}{2}$

Hollow sphere ratio $=\frac{1}{1+\frac{m R^2}{\frac{2}{3} m R^2}}=\frac{2}{5}$

Hence $ring$ correct option.

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?

Similar Questions

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

If a solid sphere of mass $1\, kg$ and radius $0.1\, m$ rolls without slipping at a uniform velocity of $1\, m/s$ along a straight line on a horizontal floor, the kinetic energy is

The moment of inertia of a body about a given axis is $2.4\ kg-m^2$. To produce a rotational kinetic energy of $750\ J$, an angular acceleration of $5\ rad/s^2$ must be applied about that axis for.......... $\sec$

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of $4\,ms ^{-1}$, is$........cm$. $\left(\right.$ take $\left.g=10\,ms ^{-2}\right)$