Non Uniform Circular Motion w/ Calculus and Vectors

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SUMMARY

This discussion focuses on solving a non-uniform circular motion problem involving a train decelerating from 90 km/hr to 50 km/hr while navigating a curve with a radius of 150 m. The key calculations involve determining both the centripetal and tangential acceleration components at the moment the train reaches 50 km/hr. The participant explores vector calculus methods, specifically using position vectors and acceleration vectors, to derive the necessary equations for this scenario. The final solution requires converting speeds from km/hr to m/s and applying Pythagorean theorem to combine the acceleration components.

PREREQUISITES
  • Understanding of non-uniform circular motion
  • Familiarity with vector calculus
  • Knowledge of centripetal and tangential acceleration
  • Ability to convert units from km/hr to m/s
NEXT STEPS
  • Study the derivation of centripetal acceleration formulas
  • Learn about tangential acceleration in circular motion
  • Explore vector calculus applications in physics problems
  • Practice converting between different speed units for motion problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in applying calculus to solve problems related to circular motion and acceleration analysis.

rue7
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Hello! I have a problem which is solvable using simpler methods, but I'm trying to use it as a bridge to understanding how to do these problems in a more rigorous setting.

Homework Statement


A train slows down as it rounds a sharp horizontal turn, slowing from 90 km/hr to 50 km/hr in the 15 s that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50 km/hr. Assume it continues to slow down at this time at the same rate.

Homework Equations





The Attempt at a Solution



First Attempt

My first attempt involved starting with a position vector

r(t) = acos(t)i + asin(t)j

where a = the radius of the curve.

However, when I took the derivative of this my attempt quickly crumbled, because:

r'(t) = v(t) = -asin(t)i + acos(t)j

in which:

|v(t)| = \sqrt{a^{2}sin^{2}(t) + a^{2}cos^{2}(t)} = a

The magnitude of the velocity vector = a at all times t? I knew this couldn't be true so I made my second attempt.

Second Attempt

This time I started with a tangent acceleration + radial acceleration vector:

a(t) = a_{t} T + a_{n}\kappa N

where T is the tangent unit vector, N is the normal (or radial) unit vector and \kappa is the curvature of the circle.

My failure here is either a lack of understanding in how to present the circular motion's acceleration or in my own ability to manipulate that vector back into a velocity vector and as well as a position vector.

Additional note:

As I said in the beginning, I know this problem is solvable by using formulas already derived using vectors and Calculus. In fact, this problem was taken from a book for a course that only requires Calculus 1. However, when I discovered how to solve Two-Dimensional (not rotational) Motion problems using derivatives and integrals rather than the "suvat" Equations of Motion beginning Physics students are taught to memorize, I immediately became curious how this can be done with Circular Motion Problems.


Many thanks if you read this far. Infinite thanks in advance for anyone who chooses to help me understand.
 
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rue7 said:
Hello! I have a problem which is solvable using simpler methods, but I'm trying to use it as a bridge to understanding how to do these problems in a more rigorous setting.

Homework Statement


A train slows down as it rounds a sharp horizontal turn, slowing from 90 km/hr to 50 km/hr in the 15 s that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50 km/hr. Assume it continues to slow down at this time at the same rate.

Homework Equations





The Attempt at a Solution



First Attempt

My first attempt involved starting with a position vector

r(t) = acos(t)i + asin(t)j

where a = the radius of the curve.

However, when I took the derivative of this my attempt quickly crumbled, because:

r'(t) = v(t) = -asin(t)i + acos(t)j

in which:

|v(t)| = \sqrt{a^{2}sin^{2}(t) + a^{2}cos^{2}(t)} = a

The magnitude of the velocity vector = a at all times t? I knew this couldn't be true so I made my second attempt.

Second Attempt

This time I started with a tangent acceleration + radial acceleration vector:

a(t) = a_{t} T + a_{n}\kappa N

where T is the tangent unit vector, N is the normal (or radial) unit vector and \kappa is the curvature of the circle.

My failure here is either a lack of understanding in how to present the circular motion's acceleration or in my own ability to manipulate that vector back into a velocity vector and as well as a position vector.

Additional note:

As I said in the beginning, I know this problem is solvable by using formulas already derived using vectors and Calculus. In fact, this problem was taken from a book for a course that only requires Calculus 1. However, when I discovered how to solve Two-Dimensional (not rotational) Motion problems using derivatives and integrals rather than the "suvat" Equations of Motion beginning Physics students are taught to memorize, I immediately became curious how this can be done with Circular Motion Problems.


Many thanks if you read this far. Infinite thanks in advance for anyone who chooses to help me understand.

The net acceleration here has two components. The centripetal acceleration, due to the fact that is is traveling at 50 km/h in a circular path of radius 150m, and a "circumferential" or tangential component due to the fact it is slowing down. It is slowing down at a rate that will cause its speed to reduce from 90 km/h to 50 km/h in 15 seconds. Calculate each component, then combine them - making use of pythagorus and trignometry as appropriate.

During you calculations you will need to change those km/h speeds to m/s.

btw the bit "Assume it continues to slow down at this time at the same rate." is there lest you think that the train, having slowed to 50km/h, then continued at that constant speed - which would have made the tangential acceleration impossible to estimate.
 

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