A wheel of radius R with an axle of radius R/ | vedclass

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(B) The torque $	au$ is given by $	au = rF \sin 	heta$,where $r$ is the distance from the axis of rotation and $	heta$ is the angle between the po

Vedclass · Accepted Answer

$3.2$

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(B) The torque $	au$ is given by $	au = rF \sin 	heta$,where $r$ is the distance from the axis of rotation and $	heta$ is the angle between the position vector and the force vector.$1$. For the force $F$ acting on the rim at an angle of $45^{\circ}$ to the tangent: The perpendicular distance from the center is $R$. The torque is $	au_1 = F \cdot R \cos(45^{\circ}) = F \cdot R \cdot \frac{1}{\sqrt{2}} \approx 0.707 \ FR$.$2$. For the force $F$ acting on the axle of radius $R/2$: The force is applied tangentially,so the torque is $	au_2 = F \cdot (R/2) = 0.5 \ FR$.$3$. For the force $2F$ acting on the rim of radius $R$: The force is applied tangentially,so the torque is $	au_3 = 2F \cdot R = 2.0 \ FR$.All these torques act in the same rotational sense (counter-clockwise).Net torque $	au_{net} = 	au_1 + 	au_2 + 	au_3 = 0.707 \ FR + 0.5 \ FR + 2.0 \ FR = 3.207 \ FR$.Thus,the magnitude of the net torque is nearly $3.2 \ FR$.

$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)

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