Rotational Motion - Merry Go Round Slowing Down | Physics with Professor Matt Anderson | M12-03

Physics with Professor Matt Anderson
1 Nov 202110:40
EducationalLearning
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TLDRIn this engaging lecture, Professor Anderson delves into the principles of rotational motion, using the example of a merry-go-round to illustrate key concepts. The discussion begins with defining an object's position in circular motion through polar coordinates (r, ΞΈ), emphasizing that r remains constant over time. The professor then introduces the concept of speed in rotational motion, calculated as the distance traveled over time, which for a full rotation is 2Ο€r/t. Applying this to a merry-go-round scenario, the speed is determined using the radius and period of rotation. With a radius of 5 meters and a period of 10 seconds, the calculated speed is Ο€ meters per second. The lecture further explores the scenario where the merry-go-round slows down over 50 seconds, using kinematic equations adapted for rotational motion to find the distance traveled and the number of rotations completed. The final calculation reveals approximately 2.5 revolutions before coming to a stop, providing a clear understanding of rotational motion and its practical applications.

Takeaways
  • πŸ“ **Polar Coordinates**: When dealing with circular motion, it's more convenient to use polar coordinates (r, ΞΈ) where r remains constant over time.
  • β­• **Circumference and Rotation**: The distance covered in one full rotation is the circumference of the circle, which is 2Ο€r.
  • πŸš€ **Speed in Rotation**: The speed of an object in rotational motion is the distance traveled over time, calculated as 2Ο€r/t.
  • 🎒 **Merry-Go-Round Example**: To find the speed on a merry-go-round, use the radius and period to calculate the initial speed (2Ο€r/t).
  • πŸ”’ **Calculating Speed**: Given a period of 10 seconds and a radius of 5 meters, the initial speed is Ο€ meters per second (approximately 3.14 m/s).
  • 🏞️ **Slowing Down Scenario**: To find out how far the merry-go-round rotates as it slows down to a stop, consider the deceleration over a given time period.
  • πŸ”΅ **Kinematic Equations Adaptation**: Use kinematic equations adapted for rotational motion to find the arc length traveled during deceleration.
  • πŸ›‘ **Deceleration to Stop**: When an object slows to a stop, the final velocity is zero, and the distance traveled can be calculated using the average velocity.
  • πŸ“ **Average Velocity Formula**: The distance traveled during deceleration is given by (1/2) * initial velocity * time, resulting in 25Ο€ meters for the given example.
  • πŸ”„ **Total Rotations**: The total number of rotations can be found by dividing the total distance traveled by the distance of one rotation (2.5 revolutions in the example).
  • ❓ **Further Questions**: Encouragement is given for students to ask questions or seek clarification if the concepts are not clear, emphasizing the importance of understanding the material.
Q & A
  • What is the main topic of the lecture?

    -The main topic of the lecture is rotational motion, with a focus on the specific example of a merry-go-round.

  • How are the position of an object in circular motion typically defined?

    -The position of an object in circular motion is typically defined using polar coordinates, which are r (the radius) and ΞΈ (theta), where r remains constant as a function of time.

  • What is the formula for the circumference of a circle?

    -The formula for the circumference of a circle is 2Ο€r, where r is the radius of the circle.

  • How is the speed of an object in circular motion calculated?

    -The speed of an object in circular motion is calculated as the distance traveled over time, which can be expressed as (2Ο€r)/t, where t is the period of rotation.

  • What is the typical period of a merry-go-round?

    -The typical period of a merry-go-round, as mentioned in the lecture, might be around 10 seconds for one complete rotation.

  • What is the initial speed of a person on the edge of a merry-go-round with a radius of 5 meters and a period of 10 seconds?

    -The initial speed of a person on the edge of the merry-go-round is Ο€ meters per second, which is approximately 3.14 meters per second.

  • If a merry-go-round slows down to a stop over 50 seconds, how can we calculate the distance it rotates?

    -We can use the kinematic equation that relates the final and initial velocities with time and acceleration to calculate the distance. Since the final velocity is zero, we use the formula s = (v_initial + v_final)/2 * t, where v_initial is the initial velocity and t is the time taken to stop.

  • What is the acceleration in the kinematic equation for the deceleration of the merry-go-round?

    -The acceleration in the equation is not directly given, but it is implied to be a constant deceleration that brings the merry-go-round to a stop over the specified time period.

  • How many rotations will the merry-go-round complete before coming to a stop if it decelerates over 50 seconds?

    -The merry-go-round will complete two and a half revolutions before coming to a stop.

  • What kinematic equation is used to calculate the distance traveled during constant acceleration or deceleration?

    -The kinematic equation used is s = s_initial + v_initial * t + 0.5 * a * t^2, where s is the final position, s_initial is the initial position, v_initial is the initial velocity, a is the acceleration, and t is the time.

  • How does the formula for calculating the distance traveled during constant acceleration or deceleration differ when applied to rotational motion?

    -In rotational motion, we use the arc length (s) instead of linear displacement (x), and the formula becomes s_final = s_initial + v_initial * t + 0.5 * a * t^2, where a is the angular acceleration or deceleration.

  • What is the significance of using polar coordinates (r and ΞΈ) for describing circular motion?

    -Polar coordinates are significant for circular motion because they simplify the description of an object's position. The radial distance (r) remains constant for an object moving at a constant radius, making ΞΈ the only variable that changes, which simplifies calculations and descriptions of circular motion.

Outlines
00:00
πŸ“ Introduction to Rotational Motion and Merry-Go-Round Example

Professor Anderson introduces the concept of rotational motion, focusing on a merry-go-round as a practical example. He explains the use of polar coordinates (r and theta) to define an object's position in circular motion, emphasizing that r remains constant over time. The professor then covers the concept of the circumference of a circle (2 pi r) and how it relates to the speed of an object completing one rotation in a given time period (s = 2 pi r / t). Using a hypothetical merry-go-round with a 10-second period and a 10-meter diameter, he calculates the initial speed of a person on the edge as approximately 3.14 meters per second. The paragraph concludes with a setup for calculating the distance the merry-go-round will rotate as it slows down over 50 seconds.

05:03
πŸ”’ Applying Kinematic Equations to Rotational Motion

The professor transitions into using kinematic equations to further analyze the rotational motion of the merry-go-round. Initially, he attempts to apply a standard kinematic equation (s = s0 + v0*t + 0.5*a*t^2) but realizes it's not suitable due to unknown acceleration. Instead, he opts for an alternative kinematic equation that uses the average velocity (s = s0 + 0.5*(v0 + vf)*t), which is applicable for constant acceleration or deceleration. With the initial velocity (v0) known from the previous calculation and the final velocity (vf) being zero (as the merry-go-round comes to a stop), he calculates the distance traveled during the 50 seconds of deceleration to be 25*pi meters. The summary concludes with the calculation of the number of rotations this distance equates to, given the radius of the merry-go-round.

10:04
πŸ”„ Conclusion on Merry-Go-Round Rotations and Final Thoughts

The final paragraph wraps up the discussion on the merry-go-round problem by determining the number of rotations made during the deceleration phase. By dividing the total distance traveled (25*pi meters) by the circumference of the merry-go-round (2*pi*5 meters), the professor calculates that the merry-go-round completes two and a half revolutions before coming to a stop. The paragraph emphasizes the reliance on kinematic equations adapted for rotational motion and invites students to seek clarification if needed. It concludes with an open invitation to the professor's office for further discussion.

Mindmap
Keywords
πŸ’‘Rotational Motion
Rotational motion refers to the movement of an object around a fixed axis. In the video, this concept is central as it discusses the motion of an object, such as a person on a merry-go-round, moving in a circular path. The script uses this to introduce polar coordinates (r and theta) as a more intuitive way to describe the position and movement of such an object.
πŸ’‘Polar Coordinates
Polar coordinates are a two-dimensional coordinate system in which each point is determined by a radius (r) from a fixed origin point and an angle (theta) from a fixed direction. The video explains that polar coordinates are particularly useful for describing circular motion, as the radius remains constant while the angle changes over time.
πŸ’‘Circumference
The circumference of a circle is the total length of the edge of the circle, which is calculated as 2 * pi * r, where r is the radius of the circle. In the context of the video, the circumference is used to calculate the distance covered by an object in one full rotation around a circle, such as the edge of a merry-go-round.
πŸ’‘Speed
Speed is a measure of how fast an object is moving, calculated as the distance traveled per unit of time. The video uses the formula for speed (distance/time) to determine the initial speed of an object in rotational motion, specifically a person on a merry-go-round, by dividing the circumference of the circle by the period of rotation.
πŸ’‘Merry-Go-Round
A merry-go-round is a ride consisting of a rotating platform with seats for people to sit on. In the video, the merry-go-round serves as a practical example to illustrate the concepts of rotational motion, speed, and kinematic equations. The script uses the merry-go-round to explore how to calculate the speed and distance traveled during deceleration.
πŸ’‘Radius
The radius is the distance from the center of a circle to any point on its edge. In the context of the video, the radius is used to calculate the circumference of the merry-go-round and, subsequently, the speed and distance traveled by an object in rotational motion. The script mentions a merry-go-round with a diameter of 10 meters, which implies a radius of 5 meters.
πŸ’‘Period
The period of an object in rotational motion is the time it takes for the object to complete one full rotation. The video uses the period, given as 10 seconds for a typical merry-go-round, to calculate the speed of an object in rotational motion and to determine how many rotations occur during a given time frame as the object slows down.
πŸ’‘Deceleration
Deceleration is the rate at which an object's speed decreases over time. The video discusses a scenario where a merry-go-round slows down to a stop, using this concept to explore how far the object will rotate during the deceleration phase. Deceleration is key to understanding the change in speed and the distance covered as the object comes to rest.
πŸ’‘Kinematic Equations
Kinematic equations are mathematical formulas used to describe the motion of an object. The video references these equations, which are typically used in linear motion, and adapts them for rotational motion to solve for the distance traveled and the number of rotations. The script uses the kinematic equation that relates initial velocity, time, and acceleration to find the arc length traveled during deceleration.
πŸ’‘Arc Length
Arc length is the distance along the path of a curve or circle. In the video, arc length is used to calculate the distance traveled by an object on the edge of a rotating merry-go-round as it slows down. The script uses the concept of arc length to determine how far the object will rotate before coming to a stop, given its initial speed and the time it takes to decelerate.
πŸ’‘Acceleration
Acceleration is the rate at which an object's velocity changes over time. Although the video does not provide the exact acceleration of the deceleration, it is implied to be a constant value that causes the merry-go-round to slow down to a stop. The concept of acceleration is important for understanding the kinematic equations used to calculate the distance traveled during the deceleration phase.
Highlights

Professor Anderson introduces the topic of rotational motion.

Polar coordinates (r and theta) are used to define an object's position in circular motion.

The distance covered in one full rotation is the circumference of the circle, which is 2Ο€r.

Speed in rotational motion is calculated as distance over time, given by 2Ο€r/t.

A specific example of a merry-go-round is used to illustrate the concepts.

The radius and period of the merry-go-round are given to calculate the speed of an object on it.

An example uses a 10-second period and a 10-meter diameter merry-go-round to calculate speed.

The initial speed of an object on the edge of the merry-go-round is calculated to be Ο€ meters per second.

A complication is added where the merry-go-round slows down to a stop over a period of 50 seconds.

Kinematic equations are applied to rotational motion to determine the distance and number of rotations.

The final position is calculated using the average velocity and time taken to slow down.

The distance the merry-go-round rotates as it slows down is calculated to be 25Ο€ meters.

The number of rotations completed as the merry-go-round comes to a stop is determined to be two and a half.

The kinematic equations are adapted for rotational motion using arc length and angular velocity.

The lecture emphasizes the practical application of kinematic equations in real-world scenarios like a merry-go-round.

The importance of understanding both linear and rotational kinematics for solving physics problems is highlighted.

An invitation is extended for students to seek clarification or further discussion during office hours.

Transcripts
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