Part of the tuning arrangement of a radio con | vedclass

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(B) The two forces form a couple. The torque $	au$ exerted by the couple is given by $	au = F 	imes d$, where $F = 0.1\ N$ is the magnitude of each

Vedclass · Accepted Answer

$0.031$

Vedclass · Answer

(B) The two forces form a couple. The torque $	au$ exerted by the couple is given by $	au = F 	imes d$, where $F = 0.1\ N$ is the magnitude of each force and $d = 0.05\ m$ is the perpendicular distance between the forces (the diameter of the wheel).$	au = 0.1\ N 	imes 0.05\ m = 0.005\ N\cdot m$.The work done $W$ when the wheel rotates by an angle $	heta$ is given by $W = 	au 	heta$.For one complete rotation, $	heta = 2\pi\ 	ext{radians}$.$W = 0.005\ N\cdot m 	imes 2\pi\ 	ext{rad} = 0.01\pi\ J$.Using $\pi \approx 3.14$, $W = 0.01 	imes 3.14 = 0.0314\ J$.Thus, the work done is about $0.031\ J$.

Part of the tuning arrangement of a radio consists of a wheel which is acted on by two parallel constant forces as shown in the figure. If the wheel rotates just once, the work done will be about (diameter of the wheel = $0.05\ m$). (in $J$)

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