r(0) Problem Statement: A toy car weighing 0.76 Ib has been released at the top of the track. The curved portion of the track is described by the equation r(0) = 7(1 – cos(0)) ft. When the car reaches a position of 0 = 269°, peed is 23.8 ft/s, and the radius of curvature of the track is 6.657 ft. Neglecting the car's size and friction, determine the following for the instant in question. ip: The track is drawn to scale. Consider printing the figure, drawing the car at your value of 0, then drawing the unit vectors accordingly. You can enlarge the figure by clicking on it. The answers depen roper definition of the unit vectors. The angle = 2.347rad The angle o = 2.330rad (1.e. the angle between gravity and tangent; it will be obtuse if the car has already passed the bottom of the track.) The normal component of acceleration: 85.0893796ft/s The tangential component of acceleration: -22.165ft/s^2 he net normal force on all wheels at the instant in question. Finally, resolve the acceleration into cylindrical components: he radial acceleration component, a, = The transverse acceleration component, ag =

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Hello, could you please work out the last three parts? I already understand how to work the first 4 parts.

r(0)
ût
Problem Statement:
A toy car weighing 0.76 Ib has been released at the top of the track. The curved portion of the track is described by the equation r(0) = 7(1 – cos(0)) ft. When the car reaches a position of 0 = 269°, its
speed is 23.8 ft/s, and the radius of curvature of the track is 6.657 ft. Neglecting the car's size and friction, determine the following for the instant in question.
Tip: The track is drawn to scale. Consider printing the figure, drawing the car at your value of 0, then drawing the unit vectors accordingly. You can enlarge the figure by clicking on it. The answers depend on
proper definition of the unit vectors.
The angle =
2.347rad
The angle o =
2.330rad
(i.e. the angle between gravity and tangent; it will be obtuse if the car has already passed the bottom of the track.)
The normal component of acceleration:
85.0893796ft/s
The tangential component of acceleration:
-22.165ft/s^2
The net normal force on all wheels at the instant in question.
Finally, resolve the acceleration into cylindrical components:
The radial acceleration component, a, =
The transverse acceleration component, ag =
Transcribed Image Text:r(0) ût Problem Statement: A toy car weighing 0.76 Ib has been released at the top of the track. The curved portion of the track is described by the equation r(0) = 7(1 – cos(0)) ft. When the car reaches a position of 0 = 269°, its speed is 23.8 ft/s, and the radius of curvature of the track is 6.657 ft. Neglecting the car's size and friction, determine the following for the instant in question. Tip: The track is drawn to scale. Consider printing the figure, drawing the car at your value of 0, then drawing the unit vectors accordingly. You can enlarge the figure by clicking on it. The answers depend on proper definition of the unit vectors. The angle = 2.347rad The angle o = 2.330rad (i.e. the angle between gravity and tangent; it will be obtuse if the car has already passed the bottom of the track.) The normal component of acceleration: 85.0893796ft/s The tangential component of acceleration: -22.165ft/s^2 The net normal force on all wheels at the instant in question. Finally, resolve the acceleration into cylindrical components: The radial acceleration component, a, = The transverse acceleration component, ag =
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