A ring of mass 10 kg and diameter 0.4 m is | vedclass

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(B) The mass of the ring is $m = 10 \ kg$ and the radius is $r = 	ext{diameter}/2 = 0.4/2 = 0.2 \ m$. The moment of inertia of a ring about its axis

Vedclass · Accepted Answer

$88$

Vedclass · Answer

(B) The mass of the ring is $m = 10 \ kg$ and the radius is $r = 	ext{diameter}/2 = 0.4/2 = 0.2 \ m$. The moment of inertia of a ring about its axis is $I = mr^2 = 10 	imes (0.2)^2 = 10 	imes 0.04 = 0.4 \ kg \cdot m^2$. The angular frequency $\omega$ is given by $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second. Given $n = 2100 \ 	ext{rpm} = 2100/60 \ 	ext{rev/s} = 35 \ 	ext{rev/s}$. Thus,$\omega = 2 	imes \pi 	imes 35 = 70\pi \ 	ext{rad/s}$. The angular momentum $L$ is given by $L = I\omega$. $L = 0.4 	imes 70\pi = 28\pi \ 	ext{kg} \cdot m^2/s$. Using $\pi \approx 22/7$,$L = 28 	imes (22/7) = 4 	imes 22 = 88 \ 	ext{kg} \cdot m^2/s$.

$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotating about its axis. If it completes $2100$ revolutions per minute,what is its angular momentum in $kg \cdot m^2/s$?

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