A circular ring of mass 10 text{ kg} rolls al | vedclass

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(D) Given, mass of the circular ring, $m = 10 	ext{ kg}$.Linear speed of the center of mass, $v = 1.5 	ext{ m/s}$.The total initial kinetic energy o

Vedclass · Accepted Answer

$-22.5$

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(D) Given, mass of the circular ring, $m = 10 	ext{ kg}$.Linear speed of the center of mass, $v = 1.5 	ext{ m/s}$.The total initial kinetic energy of the rolling ring is the sum of its rotational and translational kinetic energies:$K_i = K_{	ext{rotational}} + K_{	ext{translational}} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$.For a ring, the moment of inertia $I = m R^2$ and the angular velocity $\omega = \frac{v}{R}$.Substituting these values:$K_i = \frac{1}{2} (m R^2) \left(\frac{v}{R}ight)^2 + \frac{1}{2} m v^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2$.$K_i = 10 	imes (1.5)^2 = 10 	imes 2.25 = 22.5 	ext{ J}$.According to the work-energy theorem, the work required to stop the ring is equal to the change in kinetic energy:$W = K_f - K_i = 0 - 22.5 = -22.5 	ext{ J}$.

$A$ circular ring of mass $10 \text{ kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed of $1.5 \text{ m/s}$. The work required to stop the ring is: (in $\text{ J}$)

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