The circular disc rotates about its centre O in the direction show-n- At a certain instant point P on the rim has an acceleration a -3 -4- m - s 2- For that instanta The angular velocity - isb The angular acceleration - is

a⃗=a_t î+a_r ĵ→ω^2r =3i 4ĵ ∴ω^2 r=4 ⇒ω^2 × 0.5 =4 ⇒ω^2 =8 ∴ω =2√(2)rad/s α r=3 ⇒α× 0.5 =3 ∴α = 6rad/s^2 ...

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Standard XII

Physics

Uniform Circular Motion

The circular ...

Question

The circular disc rotates about its centre O in the direction show’n. At a certain instant point P on the rim has an acceleration $\to a = 3^i -^4^j m / s^{2}$ . For that instant
(a) The angular velocity $ω$ is _.
(b) The angular acceleration $\propto$ is _.

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Solution

$\to a = a_{t}^i + a_{r}^j \to ω^{2} r$
$= 3 i - 4^j$
$∴ ω^{2} r = 4 \Rightarrow ω^{2} \times 0.5 = 4$
$\Rightarrow ω^{2} = 8 ∴ ω = 2 \sqrt{2} r a d / s$
$α r = 3 \Rightarrow α \times 0.5 = 3 ∴ α = 6 r a d / s^{2}$

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The circular disc rotates about its centre O in the direction show’n. At a certain instant point P on the rim has an acceleration →a=3^i−^4 ^jm/s2 . For that instant (a) The angular velocity ω is ___. (b) The angular acceleration ∝ is ___.

→a=at ^i+ar^j→ω2r =3i−4^j ∴ω2r=4⇒ω2×0.5=4 ⇒ω2=8 ∴ω=2√2rad/s αr=3⇒α×0.5=3 ∴α=6rad/s2

The circular disc rotates about its centre O in the direction show’n. At a certain instant point P on the rim has an acceleration $\to a = 3^i -^4^j m / s^{2}$ . For that instant
(a) The angular velocity $ω$ is _.
(b) The angular acceleration $\propto$ is _.

$\to a = a_{t}^i + a_{r}^j \to ω^{2} r$
$= 3 i - 4^j$
$∴ ω^{2} r = 4 \Rightarrow ω^{2} \times 0.5 = 4$
$\Rightarrow ω^{2} = 8 ∴ ω = 2 \sqrt{2} r a d / s$
$α r = 3 \Rightarrow α \times 0.5 = 3 ∴ α = 6 r a d / s^{2}$