Question
The roller coaster is travelling along a track which is in the form of a cylindrical helix of 5*041
m radius rising 3 m for each half turn as shown in the figure. It is defined by θ =2t+t^3 rad,
and z = (5t – 5.8t^2) m, where t is in seconds. Determine the magnitudes of the velocity and
acceleration of the box at the instant aθ = -1.295 m/s2.
m radius rising 3 m for each half turn as shown in the figure. It is defined by θ =2t+t^3 rad,
and z = (5t – 5.8t^2) m, where t is in seconds. Determine the magnitudes of the velocity and
acceleration of the box at the instant aθ = -1.295 m/s2.
Answers
GPT 3.5
To determine the magnitudes of the velocity and acceleration of the box at the instant aθ = -1.295 m/s2, we first need to calculate the velocity and acceleration vectors at that specific instant.
Given:
θ = 2t + t^3 rad
z = 5t - 5.8t^2 m
To find the velocity vector, we need to calculate the first derivatives of the above equations with respect to time t:
v = dr/dt = (dr/dθ)(dθ/dt) + (dr/dz)(dz/dt)
Where:
dr/dθ = r'(θ) = (5.041)m
dθ/dt = 2 + 3t^2
dr/dz = z'(z) = 5 - 11.6t
dz/dt = 5 - 11.6(2 + 3t^2) = 5 - 23.2 - 34.8t^2
Now, substitute the values into the formula to get the velocity vector v:
v = (5.041)(2 + 3t^2) + (5 - 11.6t)(5 - 23.2 - 34.8t^2)
Next, differentiate v with respect to time to obtain the acceleration vector a:
a = dv/dt = (d^2r/dt^2) = ((d^2r/dθ^2)(dθ/dt)^2 + (d^2r/dz^2)(dz/dt)^2 + 2(dr/dθ)(d^2r/dz)(dθ/dt)(dz/dt))
You'd need to make the substitutions using the previously calculated values for the derivatives of r, θ, and z.
Finally, plug in the value aθ = -1.295 m/s2 into the expression for the magnitude of acceleration vector a to find the desired result.
Given:
θ = 2t + t^3 rad
z = 5t - 5.8t^2 m
To find the velocity vector, we need to calculate the first derivatives of the above equations with respect to time t:
v = dr/dt = (dr/dθ)(dθ/dt) + (dr/dz)(dz/dt)
Where:
dr/dθ = r'(θ) = (5.041)m
dθ/dt = 2 + 3t^2
dr/dz = z'(z) = 5 - 11.6t
dz/dt = 5 - 11.6(2 + 3t^2) = 5 - 23.2 - 34.8t^2
Now, substitute the values into the formula to get the velocity vector v:
v = (5.041)(2 + 3t^2) + (5 - 11.6t)(5 - 23.2 - 34.8t^2)
Next, differentiate v with respect to time to obtain the acceleration vector a:
a = dv/dt = (d^2r/dt^2) = ((d^2r/dθ^2)(dθ/dt)^2 + (d^2r/dz^2)(dz/dt)^2 + 2(dr/dθ)(d^2r/dz)(dθ/dt)(dz/dt))
You'd need to make the substitutions using the previously calculated values for the derivatives of r, θ, and z.
Finally, plug in the value aθ = -1.295 m/s2 into the expression for the magnitude of acceleration vector a to find the desired result.