A particle is undergoing circular motion of radius R with angular speed - It is brought to rest in T time and then again accelerated to the same angular speed in T2 time- The angular displacement of the particle in this time interval of 3T2is consider uniform acceleration and retardation

By using equation: 0=ω +α _1T ⇒α _1= ωT Now θ_1 =ω T+12 ( ωT )T^2 ⇒θ _1=12ω T And nbsp; ω = 0+α _2 ( T2 ) ⇒α _2=2ωT Now, θ_2 =0+12 ( 2ωT ) ( T2 )^2 = ...

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Standard XII

Physics

Average Velocity in 2D Motion

A particle is...

Question

A particle is undergoing circular motion of radius $R$ with angular speed $ω$ . It is brought to rest in $T$ time and then again accelerated to the same angular speed in $\frac{T}{2}$ time. The angular displacement of the particle in this time interval of $\frac{3 T}{2}$ is (consider uniform acceleration and retardation)
$\frac{ω T}{4}$
$\frac{ω T}{2}$
$ω T$
$\frac{3 ω T}{4}$

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Solution

The correct option is D $\frac{3 ω T}{4}$
By using equation:

$0 = ω + α_{1} T$

$\Rightarrow α_{1} = - \frac{ω}{T}$

Now $θ_{1} = ω T + \frac{1}{2} (\frac{- ω}{T}) T^{2}$

$\Rightarrow θ_{1} = \frac{1}{2} ω T$

And $ω = 0 + α_{2} (\frac{T}{2})$

$\Rightarrow α_{2} = \frac{2 ω}{T}$

Now, $θ_{2} = 0 + \frac{1}{2} (\frac{2 ω}{T}) {(\frac{T}{2})}^{2}$

$= \frac{ω T}{4}$

$\Rightarrow θ = ω T (\frac{1}{2} + \frac{1}{4}) = \frac{3 ω T}{4}$

Hence, Option $(C)$ is the correct answer.

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The correct option is D 3ωT4By using equation: 0=ω+α1T ⇒α1=−ωT Now θ1=ωT+12(−ωT)T2 ⇒θ1=12ωT And ω=0+α2(T2) ⇒α2=2ωT Now, θ2=0+12(2ωT)(T2)2 =ωT4 ⇒θ=ωT(12+14)=3ωT4 Hence, Option (C) is the correct answer.