Question

A particle is in a circular motion with radius R=0.10 m. The angular displacement is θ = 2 +4t3. The unit of θ is rad, and the unit of t is． (1) At t ＝2.0, calculate the tangential acceleration and normal acceleration． (2) When the magnitude of the tangential acceleration half of the total acceleration, calculate θ . (3) When is the magnitude of the tangential acceleration equal to normal acceleration？

Answer 1

(1) At t = 2.0 s tangential acceleration, a₁ = 4.8 m/s² and normal acceleration , a₂ = 230.4 m/s².

(2)15.8°

(3)0.55 s

Step-by-step explanation:

θ = 2 + 4t³

(1) the tangential acceleration a₁ = rα where α = angular acceleration = d²θ/dt² evaluated at t = 2.0 and r = 0.10 m

α = d²θ/dt² = 24t at t = 2 s

α = 24t = 24 × 2 = 48 rad/s²

a₁ = rα = 24tr = 0.10 m × 48 rad/s² = 4.8 m/s²

The normal acceleration a₂ = rω² where ω = angular velocity = dθ/dt = 12t² at t = 2.0 s

ω = 12t² = 12 × 2² = 48 rad/s

a₂ = rω² = 144rt⁴ = 0.1 × 48² = 230.4 m/s²

(2) The total acceleration a = √a₁² + a₂² = √[(24tr)² + (12t⁴r)²] = √[576t²r² + 144t⁸r⁴]. Since a₁ = a/2 ⇒ 24tr = √[576t²r² + 144t⁸r²]/2

48tr = √[576t²r² + 144t⁸r⁴]

squaring both sides

2304t²r² = 576t²r² + 144t⁸r²

2304t²r² - 576t²r² = 144t⁸r²

1728t²r² = 144t⁸r²

1728t²/144 = t⁸

t⁸ = 12t²

t⁸ - 12t² = 0

t²(t⁶ - 12) = 0

t² = 0 or t⁶ - 12 = 0

t = 0 or t⁶ = 12

`t = \sqrt[6]{12}\\ t = 1.51 s`

The tangential acceleration equals half the total acceleration at t = 1.51 s.

θ = 2 + 4t³ = θ = 2 + 4(1.51)³ = 15.77° ≅ 15.8°

(3) We find when a₁ = a₂

24tr = 144rt⁴

24t = 144t⁴

t⁴ = t/6

t⁴ - t/6 = 0

t(t³ - 1/6) = 0

t = 0 or t³ - 1/6 = 0

t = 0 or t = 1/∛6 = 0.55 s

So the tangential acceleration equals the normal acceleration at t = 0.55 s