Section 5.6 Circular Motion Bootcamp
Subsection 5.6.1 Position and Displacement on a Circle
Problem 5.6.1. Position, and Displacement on a Circle.
Follow the link: Checkpoint 5.1.7.
Problem 5.6.2. Rolling a Cylinder by Pushing on the Axle (JEE, 2020).
Follow the link: Checkpoint 5.1.9.
Problem 5.6.3. Arc Length, Arc Angle, Average Speed, Average Angular Speed, and Average Velocity.
Follow the link: Checkpoint 5.2.6.
Subsection 5.6.2 Velocity on a Circle
Problem 5.6.4. Angular Speed from Speed in Circular Motion.
Follow the link: Checkpoint 5.2.4.
Problem 5.6.5. (Calculus) Angular Velocity from Angle as a Function of Time.
Follow the link: Checkpoint 5.2.5.
Problem 5.6.6. Arc Length, Arc Angle, Average Speed, Average Angular Speed, and Average Velocity.
Follow the link: Checkpoint 5.2.6.
Problem 5.6.7. Circular Motion of a Cell in a Centrifuge.
Follow the link: Checkpoint 5.2.8.
Problem 5.6.8. Practice with a Friend: Regular and Angular Velocity of Earth.
Follow the link: Checkpoint 5.2.9.
Subsection 5.6.3 Uniform Circular Motion
Problem 5.6.9. Average Speed of Motion in Circular Motion and Deciding if Steady Motion.
Follow the link: Checkpoint 5.3.2.
Problem 5.6.10. Angle Covered and Distance Moved in a Uniform Circular Motion.
Follow the link: Checkpoint 5.3.3.
Problem 5.6.11. Acceleration of a Uniform Circular Motion.
Follow the link: Checkpoint 5.3.6.
Problem 5.6.12. Centripetal Acceleration in a Circular Motion.
Follow the link: Checkpoint 5.3.8.
Problem 5.6.13. Acceleration at Different Locations in an Oval Track.
Follow the link: Checkpoint 5.3.10.
Problem 5.6.14. (Calculus) Velocity of a Particle From Given Circle and Angular Speed.
Follow the link: Checkpoint 5.3.11.
Problem 5.6.15. (Calculus) Using Derivatives to Study Velocity and Acceleration of Circular Motion.
Follow the link: Checkpoint 5.3.12.
Subsection 5.6.4 Centripetal and Tangential Accelerations
Problem 5.6.16. Combining Centripetal and Tangential Accelerations in a Circular Motion.
Follow the link: Checkpoint 5.4.2.
Problem 5.6.17. Acceleration of a Car Rounding a Turn with an Increasing Speed.
Follow the link: Checkpoint 5.4.4.
Subsection 5.6.5 Using Polar Coordinates
Problem 5.6.18. Cartesian to Polar Coordinates.
Follow the link: Checkpoint 5.5.5.
Problem 5.6.19. Polar to Cartesian Coordinates.
Follow the link: Checkpoint 5.5.6.
Problem 5.6.20. Unit Vector \(\hat u_r\).
Follow the link: Checkpoint 5.5.7.
Problem 5.6.21. Practice with a Friend: Express Cartesian Components in terms of Polar Components.
Follow the link: Checkpoint 5.5.3.
Problem 5.6.22. Displacement Vector in Cartesian and Polar Coordinates.
Follow the link: Checkpoint 5.5.9.
Problem 5.6.23. Motion of a Person on a Rotating Platform Using Polar Coordinates.
Follow the link: Checkpoint 5.5.10.
Problem 5.6.24. Particle in Uniform Circular Motion Using Polar Coordinates.
Follow the link: Checkpoint 5.5.12.
Problem 5.6.25. (Calculus) Particle in a Non-uniform Circular Motion Using Polar Coordinates.
Follow the link: Checkpoint 5.5.13.
Problem 5.6.26. (Calculus) Motion on a Straight Line Described in Polar Coordinates.
Follow the link: Checkpoint 5.5.14.
Problem 5.6.27. Practice with a Friend: Elliptical Trajectory In Polar Coordinates.
Follow the link: Checkpoint 5.5.17.
Problem 5.6.28. (Calculus) Practice with a Friend: Spiral Trajectory with Zero Radial Acceleration.
Follow the link: Checkpoint 5.5.18.
Subsection 5.6.6 Miscellaneous
Problem 5.6.29. (Calculus) Pebble in a Tire of a Moving Car.
Figure 5.6.30 shows a pebble P stuck at the outside of a tire. As the tire rolls, the pebble's position in space changes. Suppose tire does not slide but rolls smoothly. The tire has radius \(R\) and is rolling at a constant speed \(v_0 \text{.}\) At \(t=0\text{,}\) the pebble was at the origin.

Find the expressions for (a) position, (b) velocity, and (c) acceleration vectors (i.e., magnitudes and directions) at an arbitraru instant \(t\) with respect to the coordinate system given in the figure.
(a) Introduce another coordinate system with origin at the center of the tire at all times and \(x\) and \(y\) axes parallel to the one given. Find position in that coordinate system first. Then translate that to the given coordinate system. (b) and (c): use derivatives.
(a) \(\vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j\text{,}\) (b) \(((1 -\cos\theta)\;v_0 , (-\sin\theta)v_0)\text{,}\) (c) \(\text{.}\)
(a) Let us introduce a coordinate system as shown in Figure 5.6.31. For convenience, I will define \(\theta\) that measures angle from negative \(y\) axis as shown.

Let \((x,y)\) and \((x',y')\) be the coordinates of P at this arbitrary instant \(t\) with respect to the two coordinate systems. It is immediately clear from the figure that
Therefore, position \(\vec r\) in original coordinate system will be
It magnitude is
We can indicate direction in space by angle with respect to \(x\) axis by using arctangent.
(b) Now, that we know the components of position, we can obtain components of velocity by taking derivatives. Thus,
Since the tire is not slipping, the rate at which angle \(\theta\) changes is related to the speed with which center of the tire is moving. The tire has rotated by \(R\Delta\theta\text{,}\) its center has moved \(v_0\Delta t\text{.}\) Equating them we get
This gives
Using this, we rewrite the components of velocity as
Therefore, magnitude of velocity, i.e., speed is
Note: this tells us that the speed of pebble when touching the ground is zero (since \(\theta=0\)) and when at the very top, it is \((2\;v_0)\) (since \(\theta=\pi\)). That is, speed of the pebble in \(Oxy\) coordinate system changes as it goes around.
The direction of velocity can again be given with respect to the \(x\) axis by using arctangent.
You can further simplify this expression by using double angle formulas. Try that. You should be able to express this without arctangent operation.
(c) Now, that we know the components of velocity, we can obtain components of acceleration by taking derivatives. Thus,
Replacing \(d\theta/dt\) by \(v_0/R\) as found in (b), we get
Therefore, magnitude of acceleration is
The direction with respect to \(x\) axis will again be obtained by using arctangent.
Of course, we can write \(cot(\theta)\) as \(\tan(\pi/2 - \theta)\text{.}\) Hence
This is actually direction towards the moving center of the tire. Take values of \(\theta=0, \pi/2, \pi, 3\pi/2, 2\pi\) and draw them on the corresponding \(Oxy\) plane. You will see that the direction is always towards the center of the tire as shown in Figure 5.6.32.
