Question

determine in instantaneous angular velocity and acceleration at time t equal to 2s if displacement is given by equation theta equal to 6t + 5 t square + 2t
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Answer 1

Question :

Determine the instantaneous angular velocity and acceleration at , t = 2 s , If the displacement of the particle is given by equation, θ = 6t + 5t² + 2t³.

Explanation :

Given :

• Position of the particle, θ = 6t + 5t² + 2t³
• Instant of time, t = 2 s

To find :

• Instantaneous angular velocity of the particle, ω = ?
• Acceleration of the particle, a = ?

Knowledge required :

• Instantaneous angular velocity of a particle is the change in position of the particle with respect to time.

Formula for instantaneous angular velocity :

⠀⠀⠀⠀⠀⠀⠀⠀⠀ω = dθ/dt

• If we differentiate the instantaneous angular velocity of a particle, we will get the Acceleration of that particle.

⠀⠀⠀⠀⠀⠀⠀⠀⠀a = d(ω)/dt

• Exponent rule of differentiation,

⠀⠀⠀⠀⠀⠀⠀⠀⠀d(x^n)/dx = n·x^(n - 1)

• Derivative of a constant term is 0.

⠀⠀⠀⠀⠀⠀⠀⠀⠀d(k)/dx = 0

Solution :

To find the instantaneous angular velocity :

By differentiating the p position of the particle w.r.t t, we get :

⠀⠀=> ω = dθ/dt

⠀⠀=> dθ/dt = d(6t + 5t² + 2t³)/dt

⠀⠀=> dθ/dt = d(6t)/dt + d(5t²)/dt + d(2t³)/dt

⠀⠀=> dθ/dt = 6t¹⁻¹ + 2 × 5t²⁻¹ + 3 × 2t³⁻¹

⠀⠀=> dθ/dt = 6t⁰ + 2 × 5t + 3 × 2t²

⠀⠀=> dθ/dt = 6 + 10t + 6t²

⠀⠀⠀⠀⠀∴ ω = 6 + 10t + 6t² m/s

Hence the instantaneous angular velocity of the particle is 6 + 10t + 6t² m/s.

Now let us find the instantaneous angular velocity of the particle at, t = 2 s.

⠀⠀=> ω = 6 + 10t + 6t²

⠀⠀=> ω_(t = 2) = 6 + 10(2) + 6(2)²

⠀⠀=> ω_(t = 2) = 6 + 20 + 6 × 4

⠀⠀=> ω_(t = 2) = 6 + 20 + 24

⠀⠀=> ω_(t = 2) = 50

⠀⠀⠀⠀⠀∴ ω = 50 m/s

Hence the instantaneous angular velocity of the particle at, t = 2 s is 50 m/s.

To find the acceleration of the particle :

By differentiating the instantaneous angular velocity of the particle w.r.t t , we get :

⠀⠀=> a = d(ω)/dt

⠀⠀=> dω/dt = d(6 + 5t + 2t²)/dt

⠀⠀=> dω/dt = d(6)/dt + d(5t)/dt + d(2t²)/dt

⠀⠀=> dω/dt = 0 + 5t¹⁻¹ + 2 × 2t²⁻¹

⠀⠀=> dω/dt = 0 + 5t⁰ + 2 × 2t¹

⠀⠀=> dω/dt = 5 + 4t

⠀⠀⠀⠀⠀∴ a = 5 + 4t m/s²

Hence the acceleration of the particle is 5 + 4t m/s².

Now let us find the acceleration of the particle at, t = 2 s.

⠀=> a = 5 + 4t

⠀=> a_(t = 2) = 5 + 4(2)

⠀=> a_(t = 2) = 5 + 8

⠀=> a_(t = 2) = 13

⠀⠀⠀⠀⠀∴ a = 13 m/s²

Hence the acceleration of the particle at, t = 2 s is 13 m/s².