A disc is rotating with angular velocity vec{ | vedclass

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(A) The power $P$ delivered by a force $\vec{F}$ acting on a particle moving with velocity $\vec{v}$ is given by $P = \vec{F} \cdot \vec{v}$.For a rot

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$(\vec{r} 	imes \vec{F}) \cdot \vec{\omega}$

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(A) The power $P$ delivered by a force $\vec{F}$ acting on a particle moving with velocity $\vec{v}$ is given by $P = \vec{F} \cdot \vec{v}$.For a rotating body,the velocity of a point with position vector $\vec{r}$ relative to the axis of rotation is $\vec{v} = \vec{\omega} 	imes \vec{r}$.Substituting this into the power formula: $P = \vec{F} \cdot (\vec{\omega} 	imes \vec{r})$.Using the property of the scalar triple product,$\vec{A} \cdot (\vec{B} 	imes \vec{C}) = (\vec{A} 	imes \vec{B}) \cdot \vec{C}$,we can rewrite the expression:$P = (\vec{F} 	imes \vec{\omega}) \cdot \vec{r} = \vec{r} \cdot (\vec{F} 	imes \vec{\omega})$.Alternatively,using the cyclic property of the scalar triple product $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$,we have:$P = \vec{F} \cdot (\vec{\omega} 	imes \vec{r}) = [\vec{F} \vec{\omega} \vec{r}] = [\vec{\omega} \vec{r} \vec{F}] = \vec{\omega} \cdot (\vec{r} 	imes \vec{F})$.Since the torque $\vec{	au} = \vec{r} 	imes \vec{F}$,the power is $P = \vec{	au} \cdot \vec{\omega} = (\vec{r} 	imes \vec{F}) \cdot \vec{\omega}$.

$A$ disc is rotating with angular velocity $\vec{\omega}$. $A$ force $\vec{F}$ acts at a point whose position vector with respect to the axis of rotation is $\vec{r}$. The power associated with the torque due to the force is given by ..........

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