Falling Box with Pulley | Physics with Professor Matt Anderson | M12-23
TLDRIn the provided script, Professor Anderson discusses a physics problem involving a box tied to a rope that passes over a disc pulley. As the box falls, it uncoils the rope and causes the disc to rotate. The challenge is to determine the acceleration of the box as it descends. The professor outlines the givens: the mass of the box and the disc are both 'm', and the radius of the disc is 'R'. A free body diagram is used to visualize the forces acting on the box, including gravity (mg) and the tension in the rope (T). The problem is solved by applying Newton's second law and considering the torque on the disc, which is related to the tension and the disc's moment of inertia. The moment of inertia for a disc is given by one half times m times R squared. The relationship between angular acceleration (ฮฑ) and linear acceleration (a) is established through the radius (R). After algebraic manipulation, the acceleration of the box is found to be two-thirds of the gravitational acceleration (G), which is less than G, as expected. The solution demonstrates the dependency of the acceleration on the object's rotation, in this case, a solid disk.
Takeaways
- ๐ The problem involves a box tied to a rope that passes over a disc pulley, causing the disc to rotate as the box falls.
- ๐ Givens include the mass of the box (m), the mass of the disc (m), and the radius of the disc (R).
- ๐ A free body diagram is essential to visualize the forces acting on the box: gravitational force (mg down) and tension (up).
- ๐ Newton's second law is applied to the box to equate the sum of forces to mass times acceleration (F = ma).
- ๐ Torque is considered for the disc's rotation, with tension times the radius (TR) equating to the moment of inertia times angular acceleration.
- ๐ The moment of inertia for a disc is given as (1/2)mr^2, which is used in the torque equation.
- โ๏ธ Angular acceleration (ฮฑ) is related to linear acceleration (a) by the radius (R), derived from the relationship a = ฮฑR.
- ๐งฎ Unit analysis helps determine the correct relationship between ฮฑ and R, leading to the simplification T = (1/2)m * a.
- โ By isolating the tension (T) and substituting it back into the box's force equation, the acceleration (a) can be solved.
- ๐ The box's acceleration (a) is found to be two-thirds of gravitational acceleration (G), or 2/3 * g.
- ๐ซ If the tension were zero (e.g., no disc), the box's acceleration would equal gravitational acceleration (a = g).
- ๐ The final answer for the box's acceleration depends on the object it's rotating with, in this case, a solid disk.
Q & A
What is the mass of the box and the disc in the given problem?
-The mass of both the box and the disc is given as 'm'.
What is the significance of the radius 'r' in the problem?
-The radius 'r' is significant as it represents the distance from the center of the disc to the point where the tension force is applied, which is critical for calculating torque.
What is the direction considered as positive in the problem?
-The positive direction is considered to be downward, as that is the direction in which the box is moving.
According to Newton's second law, what is the equation representing the forces on the box?
-The equation representing the forces on the box is mg (downward force due to gravity) minus T (upward tension force) equals m*a (mass times acceleration).
How is torque defined in the context of this problem?
-In this problem, torque is defined as the force times the lever arm. Here, the lever arm is the radius of the disc since the force is acting at a right angle to the disc.
What is the moment of inertia of a disc?
-The moment of inertia of a disc is given by one half times the mass 'm' times the radius squared 'r^2' (1/2 m r^2).
How is the angular acceleration (ฮฑ) related to the linear acceleration (a)?
-The angular acceleration (ฮฑ) is related to the linear acceleration (a) by the radius 'r' of the disc, such that ฮฑ = a/r.
What is the final expression for the acceleration 'a' of the box?
-The final expression for the acceleration 'a' of the box is a = (2/3) * g, where 'g' is the acceleration due to gravity.
What would the acceleration of the box be if the tension in the rope was zero?
-If the tension in the rope was zero, the acceleration of the box would be 'g', the acceleration due to gravity.
What is the implication of the box's acceleration being less than the acceleration due to gravity?
-The implication of the box's acceleration being less than 'g' is that the box is falling at a rate slower than it would in free fall, due to the rotational motion of the disc it is attached to.
How does the moment of inertia affect the acceleration of the box?
-The moment of inertia of the object (in this case, a disc) affects the acceleration of the box by requiring a portion of the gravitational force to be used to accelerate the rotation of the disc, thus reducing the net force available to accelerate the box's descent.
Why is unit analysis important when relating angular acceleration to linear acceleration?
-Unit analysis is important to ensure that the relationship between angular acceleration and linear acceleration is dimensionally consistent, which helps in determining the correct form of the relationship (ฮฑ = a/r or a = ฮฑ*r).
Outlines
๐ Analyzing Box and Disc System Dynamics
Professor Anderson introduces a physics problem involving a box connected to a rope that wraps around a disc pulley. The goal is to determine the acceleration of the box as it falls. Given data includes the mass (m) of both the box and the disc, and the radius (R) of the disc. The solution process involves creating a free body diagram, applying Newton's second law, and considering the torque on the disc due to the tension in the rope. The moment of inertia for the disc is used, and the relationship between angular acceleration (ฮฑ) and linear acceleration (a) is established through the radius (R). The final step involves solving for the acceleration (a), which is found to be two-thirds of the gravitational acceleration (G), indicating a fall rate less than the free-fall acceleration.
๐งฒ Tension and Acceleration in Rotational Motion
The second paragraph delves into the implications of the calculated acceleration. It confirms that the box's acceleration is less than the acceleration due to gravity (9.8 m/sยฒ), which is logical as the box is not in free fall but is rotating with the disc. The paragraph also explores a hypothetical scenario where the tension in the rope is zero, leading to the box experiencing free-fall acceleration (a = g). The summary emphasizes that the outcome is contingent on the object with which the box is rotating, and in this case, the properties of a solid disk determine the moment of inertia and thus the acceleration. Professor Anderson encourages students to seek clarification if needed.
Mindmap
Keywords
๐กFree Body Diagram
๐กNewton's Second Law
๐กTorque
๐กMoment of Inertia
๐กAngular Acceleration
๐กLinear Acceleration
๐กTension
๐กMass
๐กRadius
๐กGravitational Force
๐กAcceleration Due to Gravity
Highlights
Professor Anderson introduces a classic physics problem involving a box tied to a rope going around a disc pulley.
The mass of the box and the disc is given as 'm'.
The radius of the disc is denoted as 'capital r'.
A free body diagram is essential for solving the problem.
Forces acting on the box include gravity (mg) and tension (T) from the rope.
Torque is used to deal with the angular acceleration of the disc.
Torque is calculated as force times the lever arm, which is the radius in this case.
The moment of inertia for a disc is given as one half times m times r squared.
Angular acceleration (Alpha) is related to linear acceleration by the radius (R).
Unit analysis helps to determine the relationship between linear acceleration and angular acceleration.
The tension T can be simplified to one half times m times a when considering the radius.
Newton's second law is applied to the box to find the acceleration (a).
The final equation mg - (1/2)m*a = ma is derived to solve for acceleration.
The acceleration of the box is found to be two-thirds of G (9.8 m/s^2).
The result shows that the box falls at a rate less than the acceleration due to gravity.
If the tension were zero, the acceleration would equal g, indicating the role of the disc.
The problem's solution depends on the object's moment of inertia, specifically for a solid disk.
The lecture concludes with an invitation for further questions or clarification.
Transcripts
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