A wheel of mass 20 ,kg and radius 30 ,cm is r | vedclass

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(A) Given: Mass of wheel $m = 20 \,kg$, Radius $R = 30 \,cm = 0.3 \,m$.Initial angular speed $\omega_0 = 80 \,rpm = \frac{80 	imes 2 \pi}{60} = \frac

Vedclass · Accepted Answer

$1.06$

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(A) Given: Mass of wheel $m = 20 \,kg$, Radius $R = 30 \,cm = 0.3 \,m$.Initial angular speed $\omega_0 = 80 \,rpm = \frac{80 	imes 2 \pi}{60} = \frac{8 \pi}{3} \,rad/s$.Angular displacement $	heta = 5 \,rev = 5 	imes 2 \pi = 10 \pi \,rad$.Final angular speed $\omega = 0$.Using the rotational kinematic equation $\omega^2 = \omega_0^2 + 2 \alpha 	heta$, we find angular acceleration $\alpha$:$\alpha = \frac{\omega^2 - \omega_0^2}{2 	heta} = \frac{0 - (8 \pi / 3)^2}{2 	imes 10 \pi} = -\frac{64 \pi^2 / 9}{20 \pi} = -\frac{16 \pi}{45} \,rad/s^2$.The retarding torque $	au$ is provided by the tangential force $F$:$	au = I \alpha = F R$.For a disc, $I = \frac{1}{2} m R^2$.Thus, $F = \frac{I \alpha}{R} = \frac{1}{2} m R \alpha$.Substituting the values: $F = \frac{1}{2} 	imes 20 	imes 0.3 	imes \left| -\frac{16 \pi}{45} ight| = 10 	imes 0.3 	imes \frac{16 \pi}{45} = 3 	imes \frac{16 \pi}{45} = \frac{16 \pi}{15} \approx 1.06 \pi \,N$.

$A$ wheel of mass $20 \,kg$ and radius $30 \,cm$ is rotating at an angular speed of $80 \,rev/min$ when the motor is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in $5 \,rev$. (in $\pi \,N$)

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