A circular disc of mass 0.41 kg and radius 1 | vedclass

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(B) The total kinetic energy $(K_T)$ of a body rolling without slipping is given by the sum of translational and rotational kinetic energy: $K_T = \fr

Vedclass · Accepted Answer

$1.23$

Vedclass · Answer

(B) The total kinetic energy $(K_T)$ of a body rolling without slipping is given by the sum of translational and rotational kinetic energy: $K_T = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a circular disc,the moment of inertia $I = \frac{1}{2}mR^2$ and the rolling condition is $v = R\omega$,so $\omega = v/R$.Substituting these into the formula: $K_T = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Given $m = 0.41 \ kg$ and $v = 2 \ m/s$:$K_T = \frac{3}{4} 	imes 0.41 	imes (2)^2 = \frac{3}{4} 	imes 0.41 	imes 4 = 3 	imes 0.41 = 1.23 \ J$.

$A$ circular disc of mass $0.41 \ kg$ and radius $10 \ m$ rolls without slipping with a velocity of $2 \ m/s$. The total kinetic energy of the disc is ....... $J$.

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