If the frequency of the rotating platform is f and the distance of a boy from the centre is r, what is the area swept out per second by the line connecting the boy to the centre?

Question

If the frequency of the rotating platform is f and the distance of a boy from the centre is r, what is the area swept out per second by the line connecting the boy to the centre? (A) π r f (B) 2 π r f (C) π r2 f (D) 2 π r2 f

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/072310df0f2f96d702260cc110f7a142-.png" /> Area of sector (A)=(1/2) r2 θ (dA/dt)=(1/2) r2 (d θ/d t)=(r2/2) ⋅ ω=(r2/2)(2 π f)=π r2 f

Q. If the frequency of the rotating platform is $f$ and the distance of a boy from the centre is $r,$ what is the area swept out per second by the line connecting the boy to the centre?

$π r f$

$2 π r f$

$π r^{2} f$

$2 π r^{2} f$

Area of sector $(A) = \frac{1}{2} r^{2} θ$
$\frac{d A}{d t} = \frac{1}{2} r^{2} \frac{d θ}{d t} = \frac{r ^{2}}{2} \cdot ω = \frac{r ^{2}}{2} (2 π f) = π r^{2} f$

Q. If the frequency of the rotating platform is f and the distance of a boy from the centre is r, what is the area swept out per second by the line connecting the boy to the centre?

πrf

2πrf

πr2f

2πr2f