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
Answer:
- The angular acceleration (α) of the particle is 12 rad / sec²
Given:
- The given Relation is θ = t³ + 2 t + 1
Explanation:
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.
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.
Answer:
Given:
Angular displacement is a function of time as follows
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:
Now , angular Velocity be :
Now finding Angular acceleration :
So putting the value of t , we get :
So final answer :