A disc rotates about its axis with a constant angular acceleration of 4 rads2- Radial acceleration of a particle at a distance of 1 cm from the axis at the end of 1 sec is-

T = 1 sec, r = 1 cm = 0.01 m α = 4 rd/s^2 Therefore angular velocity at t= 1 second, ω = α t = 4 rad/s Therefore radial acceleration, a_c = ω^2 r = 0.16 m/s ...

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Standard XII

Physics

Deducing Mathematically

A disc rotate...

Question

A disc rotates about its axis with a constant angular acceleration of $4 \frac{r a d}{s^{2}} .$ Radial acceleration of a particle at a distance of $1 c m$ from the axis at the end of $1 s e c$ is.

16 \frac{c m}{s^{2}}

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1.6 \frac{c m}{s^{2}}

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16 \frac{m}{s^{2}}

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1.6 \frac{m}{s^{2}}

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Solution

The correct option is A $16 \frac{c m}{s^{2}}$
$t$ = 1 sec, $r$ = 1 cm = 0.01 m

$α$ = 4 $\frac{r d}{s^{2}}$

Therefore angular velocity at t= 1 second,
$ω$ = $α t$ = 4 $\frac{r a d}{s}$

Therefore radial acceleration,

$a_{c}$ = $ω^{2} r$ = 0.16 $\frac{m}{s^{2}}$ = 16 $\frac{c m}{s^{2}}$

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A disc rotates about its axis with a constant angular acceleration of 4 rads2. Radial acceleration of a particle at a distance of 1 cm from the axis at the end of 1 sec is.

The correct option is A 16cms2t = 1 sec, r = 1 cm = 0.01 m α = 4 rds2 Therefore angular velocity at t= 1 second, ω = αt = 4 rads Therefore radial acceleration, ac = ω2r = 0.16 ms2 = 16 cms2

A disc rotates about its axis with a constant angular acceleration of $4 \frac{r a d}{s^{2}} .$ Radial acceleration of a particle at a distance of $1 c m$ from the axis at the end of $1 s e c$ is.

The correct option is A $16 \frac{c m}{s^{2}}$
$t$ = 1 sec, $r$ = 1 cm = 0.01 m

$α$ = 4 $\frac{r d}{s^{2}}$

Therefore angular velocity at t= 1 second,
$ω$ = $α t$ = 4 $\frac{r a d}{s}$

Therefore radial acceleration,

$a_{c}$ = $ω^{2} r$ = 0.16 $\frac{m}{s^{2}}$ = 16 $\frac{c m}{s^{2}}$