A particle starting from rest moves with the acceleration 𝑎̅ = 4𝑡𝑖̂ - 3𝑡2𝑗 - 6𝑘̂ m/s2. Determine

Question

A particle starting from rest moves with the acceleration 𝑎̅ = 4𝑡𝑖̂ - 3𝑡2𝑗 - 6𝑘̂ m/s2. Determine

the principal radius of curvature of its path at t = 1.*5*041sec

Answer 1

To find the principal radius of curvature of the path at t = 1.5041 sec, we need to first find the velocity and acceleration of the particle at that time.

Given acceleration 𝑎̅ = 4𝑡𝑖̂ - 3𝑡^2𝑗̂ - 6𝑘̂ m/s^2, the velocity of the particle at time t is given by integrating the acceleration:
v̅ = ∫a̅ dt

v̅ = ∫(4𝑡𝑖̂ - 3𝑡^2𝑗̂ - 6𝑘̂) dt
v̅ = 2𝑡^2𝑖̂ - 𝑡^3𝑗̂ - 6t𝑘̂

Now, find the velocity of the particle at t = 1.5041 sec:
v̅ = 2(1.5041)^2𝑖̂ - (1.5041)^3𝑗̂ - 6(1.5041)𝑘̂
v̅ = 2.2633𝑖̂ - 3.5911𝑗̂ - 9.0246𝑘̂ m/s

The principal radius of curvature is given by the formula:
r = |v̅|^3 / |a̅⊥|

Where |a̅⊥| is the magnitude of the component of acceleration perpendicular to the velocity, given by:
a̅⊥ = a̅ - (a̅ · v̅)v̅ / |v̅|

Now, find the perpendicular component of acceleration at t = 1.5041 sec:
a̅⊥ = a̅ - (a̅ · v̅)v̅ / |v̅|
a̅⊥ = (4𝑖̂ - 3(1.5041)𝑗̂ - 6𝑘̂) - ((4𝑖̂ - 3(1.5041)𝑗̂ - 6𝑘̂) · (2.2633𝑖̂ - 3.5911𝑗̂ - 9.0246𝑘̂))(2.2633𝑖̂ - 3.5911𝑗̂ - 9.0246𝑘̂) / |2.2633𝑖̂ - 3.5911𝑗̂ - 9.0246𝑘̂|
a̅⊥ = (4𝑖̂ - 4.5123𝑗̂ - 6𝑘̂) - ((4 - 4.5123 - 6)(4 - 4.5123 - 6))/(2.2633^2 + 3.5911^2 + 9.0246^2) m/s^2
a̅⊥ = (4𝑖̂ - 4.5123𝑗̂ - 6𝑘̂) - (-10.0915)(-10.0915)/(58.5073) m/s^2
a̅⊥ = (4𝑖̂ - 4.5123𝑗̂ - 6𝑘̂) - 101.8328/58.5073 m/s^2
a̅⊥ = (4𝑖̂ - 4.5123𝑗̂ - 6𝑘̂) - 1.7417 m/s^2
a̅⊥ = (4𝑖̂ - 4.5123𝑗̂ - 7.7417𝑘̂) m/s^2

Now, find the magnitude of velocity at t = 1.5041 sec:
|v̅| = sqrt((2.2633)^2 + (-3.5911)^2 + (-9.0246)^2)
|v̅| = sqrt(5.1205 + 12.9102 + 81.4327)
|v̅| = sqrt(99.4634)
|v̅| = 9.9731 m/s

Now, we can calculate the principal radius of curvature at t = 1.5041 sec:
r = |v̅|^3 / |a̅⊥|
r = (9.9731)^3 / sqrt((4)^2 + (-4.5123)^2 + (-7.7417)^2)
r = 993.115 / sqrt(16 + 20.3835 + 59.8347)
r = 993.115 / sqrt(96.2183)
r = 993.115 / 9.8081
r ≈ 101.13 meters

Therefore, the principal radius of curvature of the particle's path at t = 1.5041 sec is approximately 101.13 meters.