A wheel rotating at 12 revolution per second is brought to rest in 6 s. The average angular deceleration in rad/s2 of the wheel during this process is

Question

A wheel rotating at 12 revolution per second is brought to rest in 6 s. The average angular deceleration in rad/s2 of the wheel during this process is (A) 4 π  (B) 4 (C) 72 (D) 1/π

Tardigrade Admin · Accepted Answer

W10=12 × 2 π ( rad /s) =24 π ( rad /s) Wf =0 Δ t =6 s As Wf =Wi+α Δ t We have, α =(-Wi/Δ t) =-(24 π/6) =-4 π ( rad / Sec 2)

Q. A wheel rotating at $12$ revolution per second is brought to rest in $6 s$ . The average angular deceleration in $r a d / s^{2}$ of the wheel during this process is

$4 π$

4

72

$1/ π$

$π$

$W_{1}^{0} = 12 \times 2 π \frac{r a d}{s}$
$= 24 π \frac{r a d}{s}$
$W_{f} = 0$
$Δ t = 6 s$
As $W_{f} = W_{i} + α Δ t$
We have, $α = \frac{- W _{i}}{Δ t}$
$= - \frac{24 π}{6}$
$= - 4 π \frac{r a d}{S e c ^{2}}$

Q. A wheel rotating at 12 revolution per second is brought to rest in 6s. The average angular deceleration in rad/s2 of the wheel during this process is

4π

4

72

1/π

π

W10​=12×2πsrad​ =24πsrad​ Wf​=0 Δt=6s As Wf​=Wi​+αΔt We have, α=Δt−Wi​​ =−624π​ =−4πSec2rad​

Q. A wheel rotating at $12$ revolution per second is brought to rest in $6 s$ . The average angular deceleration in $r a d / s^{2}$ of the wheel during this process is

$4 π$

$1/ π$

$π$

$W_{1}^{0} = 12 \times 2 π \frac{r a d}{s}$
$= 24 π \frac{r a d}{s}$
$W_{f} = 0$
$Δ t = 6 s$
As $W_{f} = W_{i} + α Δ t$
We have, $α = \frac{- W _{i}}{Δ t}$
$= - \frac{24 π}{6}$
$= - 4 π \frac{r a d}{S e c ^{2}}$