Physics, asked by jainrishabh3309, 8 months ago

The angular displacement of a particle is given by theta = t^3 + 2t +1, where t is time in seconds. Its angular acceleration at t=2s is

Answers

Answered by ShivamKashyap08
49

Answer:

  • The angular acceleration (α) of the particle is 12 rad / sec²

Given:

  1. The given Relation is θ = t³ + 2 t + 1

Explanation:

\rule{300}{1.5}

From the relation we know,

θ = t³ + 2 t + 1

Differentiating it w.r.t Time.

⇒ d θ / d t = d (t³ + 2 t + 1) / d t

∵ [ ω = d θ / d t ]

Differentiation the equation.

⇒ ω = 3 × t⁽³ ⁻ ¹⁾ + 1 × 2 t⁽¹ ⁻ ¹⁾ + 0

⇒ ω = 3 t² + 2 t

ω = 3 t² + 2 t rad / sec.

\rule{300}{1.5}

\rule{300}{1.5}

From the above relation we know,

ω = 3 t² + 2 t

Differentiating it w.r.t Time.

⇒ d ω / d t = d (3 t² + 2) / d t

∵ [ α = d ω / d t ]

Differentiation the equation.

⇒ α = 2 × 3 t⁽² ⁻ ¹⁾ + 0

α = 6 t  

Substituting the value, of time ( t = 2 sec )

⇒ α = 6 × (2)

⇒ α = 6 × 2

⇒ α = 12

α = 12 rad /sec²

The angular acceleration (α) of the particle is 12 rad / sec².

\rule{300}{1.5}

Answered by nirman95
40

Answer:

Given:

Angular displacement is a function of time as follows

 \theta \:  =  {t}^{3}  + 2t + 1

To find:

Angular Acceleration at t = 2 sec

Concept:

Angular displacement is the rotational analogue of displacement (in linear motions). It is an axial vector directed perpendicular to the plane in which the object undergoes rotational motion.

Angular Acceleration is the rotational analogue of Linear Acceleration. It is given by the 2nd order derivative of angular displacement.

Calculation:

 \theta \:  =  {t}^{3}  + 2t + 1

Now , angular Velocity be :

 \omega =  \dfrac{d \theta}{dt}

 =  >  \omega =  \dfrac{d( {t}^{3}  + 2t + 1)}{dt}

 =  >  \omega = 3  {t}^{2}  + 2

Now finding Angular acceleration :

 \alpha  =  \dfrac{d \omega}{dt}  =  \dfrac{ {d}^{2} \theta }{d {t}^{2} }

 =  >  \alpha  =  \dfrac{d(3 {t}^{2}  + 2)}{dt}

 =  >  \alpha  = 6t

So putting the value of t , we get :

 =  >  \alpha  = 6 \times 2

 =  >  \alpha  = 12 \: rad \:  {s}^{ - 2}

So final answer :

 \boxed{ \blue{ \bold{ \huge{  \alpha  = 12 \: rad \:  {s}^{ - 2} }}}}

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