Problem Overview
We analyze the space curve kinematics, total track arc length, curvature, and acceleration decomposition for an entertainment roller coaster loop.
Step-by-Step Mathematical Solution
Step 1: Velocity, Acceleration, and Scalar Speed
State the Vector Derivative Formulas :
v β ( t ) = r β β² ( t ) = β¨ 6 cos β‘ t , β
β 6 sin β‘ t , β
β 8 β© Β m/s \vec{v}(t) = \vec{r}'(t) = \langle 6\cos t, \; 6\sin t, \; 8 \rangle\text{ m/s} v ( t ) = r β² ( t ) = β¨ 6 cos t , 6 sin t , 8 β© Β m/s a β ( t ) = v β β² ( t ) = β¨ β 6 sin β‘ t , β
β 6 cos β‘ t , β
β 0 β© Β m/s 2 \vec{a}(t) = \vec{v}'(t) = \langle -6\sin t, \; 6\cos t, \; 0 \rangle\text{ m/s}^2 a ( t ) = v β² ( t ) = β¨ β 6 sin t , 6 cos t , 0 β© Β m/s 2 (Defined in Derivatives of Vector-Valued Functions .)
State the Scalar Speed Formula :
v ( t ) = β₯ v β ( t ) β₯ = ( 6 cos β‘ t ) 2 + ( 6 sin β‘ t ) 2 + 8 2 = 36 ( cos β‘ 2 t + sin β‘ 2 t ) + 64 = 36 + 64 = 100 = 10 Β m/s v(t) = \|\vec{v}(t)\| = \sqrt{(6\cos t)^2 + (6\sin t)^2 + 8^2} = \sqrt{36(\cos^2 t + \sin^2 t) + 64} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m/s} v ( t ) = β₯ v ( t ) β₯ = ( 6 cos t ) 2 + ( 6 sin t ) 2 + 8 2 β = 36 ( cos 2 t + sin 2 t ) + 64 β = 36 + 64 β = 100 β = 10 Β m/s constant at v ( t ) = 10 Β m/s v(t) = 10\text{ m/s} v ( t ) = 10 Β m/s Step 2: Compute Total Arc Length of the Loop (0 β€ t β€ 2 Ο 0 \le t \le 2\pi 0 β€ t β€ 2 Ο
State the Definite Arc Length Formula :
s = β« a b β₯ r β β² ( t ) β₯ β d t s = \int_{a}^{b} \|\vec{r}'(t)\|\,dt s = β« a b β β₯ r β² ( t ) β₯ d t (Defined in Arc Length of a Space Curve .)
s = β« 0 2 Ο 10 β d t = [ 10 t ] 0 2 Ο = 20 Ο β 62.83 Β meters s = \int_{0}^{2\pi} 10\,dt = [10t]_{0}^{2\pi} = 20\pi \approx 62.83\text{ meters} s = β« 0 2 Ο β 10 d t = [ 10 t ] 0 2 Ο β = 20 Ο β 62.83 Β meters Step 3: Compute Curvature ΞΊ \kappa ΞΊ Ο \rho Ο
State the Cross-Product Curvature Formula :
ΞΊ = β₯ v β Γ a β β₯ β₯ v β β₯ 3 \kappa = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|^3} ΞΊ = β₯ v β₯ 3 β₯ v Γ a β₯ β (Defined in Curvature of a Space Curve .)
Calculate the cross product v β Γ a β \vec{v} \times \vec{a} v Γ a v β Γ a β = β£ i ^ j ^ k ^ 6 cos β‘ t 6 sin β‘ t 8 β 6 sin β‘ t 6 cos β‘ t 0 β£ \vec{v} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6\cos t & 6\sin t & 8 \\ -6\sin t & 6\cos t & 0 \end{vmatrix} v Γ a = β i ^ 6 cos t β 6 sin t β j ^ β 6 sin t 6 cos t β k ^ 8 0 β β v β Γ a β = β¨ 0 β 48 cos β‘ t , β
β β 48 sin β‘ t β 0 , β
β 36 cos β‘ 2 t β ( β 36 sin β‘ 2 t ) β© = β¨ β 48 cos β‘ t , β
β β 48 sin β‘ t , β
β 36 β© \vec{v} \times \vec{a} = \langle 0 - 48\cos t, \; -48\sin t - 0, \; 36\cos^2 t - (-36\sin^2 t) \rangle = \langle -48\cos t, \; -48\sin t, \; 36 \rangle v Γ a = β¨ 0 β 48 cos t , β 48 sin t β 0 , 36 cos 2 t β ( β 36 sin 2 t )β© = β¨ β 48 cos t , β 48 sin t , 36 β©
Compute magnitude:
β₯ v β Γ a β β₯ = ( β 48 cos β‘ t ) 2 + ( β 48 sin β‘ t ) 2 + 36 2 = 48 2 ( cos β‘ 2 t + sin β‘ 2 t ) + 36 2 = 2304 + 1296 = 3600 = 60 \|\vec{v} \times \vec{a}\| = \sqrt{(-48\cos t)^2 + (-48\sin t)^2 + 36^2} = \sqrt{48^2(\cos^2 t + \sin^2 t) + 36^2} = \sqrt{2304 + 1296} = \sqrt{3600} = 60 β₯ v Γ a β₯ = ( β 48 cos t ) 2 + ( β 48 sin t ) 2 + 3 6 2 β = 4 8 2 ( cos 2 t + sin 2 t ) + 3 6 2 β = 2304 + 1296 β = 3600 β = 60
Calculate curvature:
ΞΊ = 60 ( 10 ) 3 = 60 1000 = 0.06 Β m β 1 = 3 50 Β m β 1 \kappa = \frac{60}{(10)^3} = \frac{60}{1000} = 0.06\text{ m}^{-1} = \frac{3}{50}\text{ m}^{-1} ΞΊ = ( 10 ) 3 60 β = 1000 60 β = 0.06 Β m β 1 = 50 3 β Β m β 1
Radius of curvature:
Ο = 1 ΞΊ = 1 0.06 = 50 3 β 16.67 Β meters \rho = \frac{1}{\kappa} = \frac{1}{0.06} = \frac{50}{3} \approx 16.67\text{ meters} Ο = ΞΊ 1 β = 0.06 1 β = 3 50 β β 16.67 Β meters
Step 4: Tangential and Normal Acceleration Components
State the Acceleration Decomposition Formulas :
a β ( t ) = a T T ^ + a N N ^ \vec{a}(t) = a_T \hat{T} + a_N \hat{N} a ( t ) = a T β T ^ + a N β N ^ (Defined in Tangential and Normal Components of Acceleration .)
Tangential Component:
a T = v β β
a β β₯ v β β₯ = ( 6 cos β‘ t ) ( β 6 sin β‘ t ) + ( 6 sin β‘ t ) ( 6 cos β‘ t ) + ( 8 ) ( 0 ) 10 = β 36 sin β‘ t cos β‘ t + 36 sin β‘ t cos β‘ t 10 = 0 Β m/s 2 a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|} = \frac{(6\cos t)(-6\sin t) + (6\sin t)(6\cos t) + (8)(0)}{10} = \frac{-36\sin t\cos t + 36\sin t\cos t}{10} = 0\text{ m/s}^2 a T β = β₯ v β₯ v β
a β = 10 ( 6 cos t ) ( β 6 sin t ) + ( 6 sin t ) ( 6 cos t ) + ( 8 ) ( 0 ) β = 10 β 36 sin t cos t + 36 sin t cos t β = 0 Β m/s 2 (Since speed is constant, a T = d d t ( 10 ) = 0 a_T = \frac{d}{dt}(10) = 0 a T β = d t d β ( 10 ) = 0 Normal Component:
a N = ΞΊ β₯ v β β₯ 2 = ( 0.06 ) ( 10 ) 2 = ( 0.06 ) ( 100 ) = 6.0 Β m/s 2 a_N = \kappa \|\vec{v}\|^2 = (0.06)(10)^2 = (0.06)(100) = 6.0\text{ m/s}^2 a N β = ΞΊ β₯ v β₯ 2 = ( 0.06 ) ( 10 ) 2 = ( 0.06 ) ( 100 ) = 6.0 Β m/s 2
Step 5: Passenger Biomechanical Safety Assessment
Calculated centripetal acceleration: a N = 6.0 Β m/s 2 a_N = 6.0\text{ m/s}^2 a N β = 6.0 Β m/s 2
Expressed in G's: a N g = 6.0 9.81 β 0.61 Β g \frac{a_N}{g} = \frac{6.0}{9.81} \approx 0.61\text{ g} g a N β β = 9.81 6.0 β β 0.61 Β g
Allowable limit: a N , max β‘ = 1.0 Β g = 9.81 Β m/s 2 a_{N,\max} = 1.0\text{ g} = 9.81\text{ m/s}^2 a N , m a x β = 1.0 Β g = 9.81 Β m/s 2
Engineering Conclusion
Because the normal acceleration is constant at a N = 6.0 Β m/s 2 a_N = 6.0\text{ m/s}^2 a N β = 6.0 Β m/s 2 0.61 Β g 0.61\text{ g} 0.61 Β g 1.0 Β g 1.0\text{ g} 1.0 Β g Ο = 16.67 Β m \rho = 16.67\text{ m} Ο = 16.67 Β m