A robot arm moves so that p travels in a circle about point b which is not moving. knowing that p starts from rest, and its speed increases at a constant rate of 10mm/s, determine (a) the magnitude of the acceleration when t=4s, (b) the time for the magnitude of the acceleration to be 80 mm/s^2

(a)10.20 mm/s² (b) 403200 s⁴

Explanation:

Solution

Recall that,

The tangible acceleration is a₁ = 10mm/s

The speed = a₁t

Normal acceleration =  aₙ = v₂ /р = a₁²t₂/ р

where р = 0.8m = 800 mm

Now,

When t = 4s

v = (10) (4) = 40 mm/s

Thus,

aₙ = (40)² /800 = 2 mm/s²

Then

The acceleration is,

a = √a₁² + aₙ² = √ (10)² + (2)²

a = 10.20 mm/s²

(b) The time for he magnitude of the acceleration to be 80 mm/s^2

a² =  aₙ² +a₁²

(80)² + [ (10)²t²/800]² + 10²

so,

t⁴ = 403200 s⁴

answer:

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