If a disc starting from rest acquires an angular velocity of $240$ revolution/ min in $10 s$ , then its angular acceleration will be

1.52 r a d s^{- 1}

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3.11 r a d s^{- 1}

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2.51 r a d s^{- 1}

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1.13 r a d s^{- 1}

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Solution

The correct option is C $2.51 r a d s^{- 1}$
The angular acceleration, $α = \frac{ω_{2} - ω_{1}}{t}$
Here, $ω_{2} = 2 π n_{2} = 2 π \times \frac{240}{60}$
and $ω_{1} = 2 π n_{1} = 0$
$∴ α = \frac{2 π (\frac{240}{60} - 0)}{10} = 2.51 r a d s^{- 1}$ .

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If a disc starting from rest acquires an angular velocity of 240 revolution/ min in 10 s, then its angular acceleration will be

The correct option is C 2.51 rads−1The angular acceleration, α=ω2−ω1tHere, ω2=2πn2=2π×24060and ω1=2πn1=0∴α=2π(24060−0)10=2.51 rads−1.

If a disc starting from rest acquires an angular velocity of $240$ revolution/ min in $10 s$ , then its angular acceleration will be

The correct option is C $2.51 r a d s^{- 1}$
The angular acceleration, $α = \frac{ω_{2} - ω_{1}}{t}$
Here, $ω_{2} = 2 π n_{2} = 2 π \times \frac{240}{60}$
and $ω_{1} = 2 π n_{1} = 0$
$∴ α = \frac{2 π (\frac{240}{60} - 0)}{10} = 2.51 r a d s^{- 1}$ .