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Rotational Motion - 01 || Torque and Moment Of Inertia || NEET Physics Crash Course

Physics Wallah - Alakh Pandey・2 minutes read

A session on rotation covers concepts such as torque, moment of inertia, and equilibrium, with detailed explanations and practical examples. The importance of understanding formulas, calculations, and application of principles in physics is emphasized for accurate solutions and learning.

Insights

  • The session covers a range of topics including torque, moment of inertia, and pure rotation, offering a comprehensive understanding of rotational motion.
  • The significance of understanding torque and moment of inertia in rotational motion is highlighted, emphasizing their crucial roles in calculations and applications.
  • The formula for torque is detailed, focusing on the product of the perpendicular distance from the pivot point and the applied force, with practical examples provided for clarity.
  • The importance of equilibrium in bodies, both in terms of translation and rotational equilibrium, is explained, stressing the need for zero net force and torque for stability.
  • The Parallel Axis Theorem is discussed as a key concept for calculating the moment of inertia of various bodies, providing a fundamental understanding of rotational dynamics.

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Recent questions

  • What is torque in rotational motion?

    Torque is the product of perpendicular distance and force.

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Summary

00:00

"Understanding Rotation: Torque, Inertia, and Velocity"

  • The session begins with a call for everyone to join, emphasizing the start of a rotation session.
  • The challenging nature of the topic is acknowledged, promising an enjoyable learning experience.
  • The session will cover previous year questions, concepts, shortcuts, and practice sheets related to rotation.
  • The importance of understanding torque and moment of inertia in rotational motion is highlighted.
  • The concept of pure rotation is explained, emphasizing that all particles move in circles with centers on the axis of rotation.
  • The relationship between angular and linear velocities in rotational motion is discussed.
  • The concept of torque, or moment of force, is introduced, with a practical example involving a screw and force application.
  • The formula for torque is detailed as the product of the perpendicular distance from the pivot point and the applied force.
  • The significance of the normal reaction passing through the center of mass for torque calculations is explained.
  • The formula for torque is further elaborated by considering the perpendicular distance and force application point.

17:18

Understanding Torque: Formula and Calculation Methods

  • Torque can be expressed as f * r perpendicular, with r being perpendicular to the force.
  • The formula for torque can be written as r * f perpendicular or f * r perpendicular.
  • The choice between r * f perpendicular or f * r perpendicular depends on the visibility of the dependent force or the perpendicular r.
  • Torque can be calculated using the formula r cross f, where the direction of torque is determined by the cross product of r and f.
  • The direction of torque can be identified by pointing the thumb in the direction of rotation.
  • The dot product of two perpendicular vectors is zero, as seen in the formula for torque.
  • When calculating torque, the force and distance from the point of suspension are crucial.
  • The torque due to weight can be determined by considering the angle and distance from the point of suspension.
  • The torque of tension is zero when the tension force passes through the axis of rotation.
  • Net torque is the resultant of all torque forces, with clockwise torques considered negative and anticlockwise torques positive.

35:28

Understanding Torque and Equilibrium in Physics

  • Newton discussed the torque of different forces, explaining the calculations involved.
  • Torque of 8 Newtons at 30 degrees from the axis of rotation was analyzed.
  • The torque of 6 Newtons was determined to be zero as it passed through the axis of rotation.
  • The torque of 9 Newtons at a 90-degree angle was calculated to be 1.8 Newton Meters clockwise.
  • The process of finding torque through cross products was demonstrated with numerical examples.
  • The concept of equilibrium in bodies was explained, emphasizing zero net force and torque for translation and rotational equilibrium.
  • The significance of zero net torque in static equilibrium was highlighted, indicating equal clockwise and anticlockwise torques.
  • The direction of angular velocity and acceleration in rotational equilibrium was discussed.
  • The calculation of torque about any point in a system was explained to be zero in equilibrium.
  • A practical example involving balancing weights on a plank was illustrated, showcasing the application of torque calculations.

51:59

Balancing Forces and Torques in Seating Arrangement

  • The value of one's son becomes 5/3, costing 1.67.
  • A 15kg person couldn't be seated on one side due to balance issues.
  • Seating a 15kg person would require adjusting the weight on the other side.
  • The story involves deciding which child should sit where based on weight.
  • Placing the heavier child closer helps balance due to torque.
  • Increasing the distance of the lighter child helps maintain balance.
  • Four equal and parallel forces on a rod create torque and linear motion.
  • Net force is zero, but net torque is not, indicating the body isn't at rest.
  • Tension in threads supporting a 20kg rod and a 40kg mass is calculated.
  • Finding the normal reaction of supports for a 10kg rod involves balancing forces and torques.

01:07:14

Torque and Normal Reactions in Physics

  • The value of n2 is 6g and n1 is also 6g.
  • If n2 becomes 60 when jagi is 10, then n2 is 40.
  • Normal reaction will not affect you if a scale is placed on top.
  • The normal reaction will be covered with normal reaction.
  • The net torque acting on people's bodies is zero.
  • The torque point about lu a and b is discussed.
  • The torque of w * d - x is equal to the torque of a.
  • The torque of f1 + f2 is equal to the torque of f3.
  • The perpendicular distance of friction is h/2.
  • The value of x/2 is equal to the coefficient of friction.

01:26:33

Normal reactions and torque in equilibrium.

  • The normal reaction will appear on the left side of O.
  • The normal reaction will be to the left of O.
  • The normal reaction will rotate opposite to the friction.
  • The normal reaction is located on the left.
  • The torque of the normal force acting on the block is found.
  • The torque of the normal reaction is equal to the torque of friction.
  • The center of mass is where the normal reaction is located.
  • The ladder problem involves normal reactions and friction forces.
  • The ladder will not slip due to normal reactions and friction.
  • The net torque must be zero for equilibrium.

01:42:49

Forces, Torque, and Friction in Physics

  • n1 is equal to 10g, n1 = 10g of g value beta 10
  • Normal reaction is 100 Newtons, present in all three instances
  • Frictional force equation involves n2
  • Friction is equal to n2
  • Net torque is zero, calculated about point A
  • Torque of n2 is calculated, with a perpendicular distance of 2ls37
  • Naji's torque is considered, with a perpendicular distance of 12s37
  • N2 value is determined as 66.66 Newtons
  • Forces are removed when calculating net torque
  • Cyclist bends at turns, with centrifugal force causing the bend

01:58:38

Forces and Moments in Object Equilibrium

  • The force preventing an object from being thrown out is the centrifugal force.
  • When a brother turns around, one edge of an object will be down.
  • The force behind throwing an object out is the normal reaction.
  • Frictional force plays a crucial role in saving objects from being thrown out.
  • The equilibrium of an object involves the net force being zero.
  • The torque zero point is crucial for balancing forces.
  • The moment of inertia is a property of a body opposing changes in its state of rest or motion.
  • The net moment of inertia of a system of particles is the sum of individual moments.
  • The moment of inertia of a system about an axis passing through one corner of an equilateral triangle is calculated by considering the distances of masses from the axis.
  • The moment of inertia of a system about an axis perpendicular to the plane of a triangle is determined by calculating the perpendicular distances of masses from the axis.

02:16:42

NEET 2016 Physics: Center of Mass

  • NEET 2016 is mentioned, indicating a focus on a specific exam year.
  • Discussion on a light rod of length 'a' with masses 'm1' and 'm2' attached to its ends.
  • Explanation of a light rod having no mass, clarifying its characteristics.
  • Query about the center of mass of the system with masses 'm1' and 'm2'.
  • Calculation of the center of mass distance from the ends of the rod.
  • Detailed steps on finding the moment of inertia of the system.
  • Introduction to the Mentos method for reducing mass in a system.
  • Explanation of moment of inertia for continuous mass distribution.
  • Derivation of moment of inertia for a uniform linear rod.
  • Calculation of moment of inertia for a uniform circular ring.

02:35:52

Calculating Moment of Inertia with Theorems

  • Moment of inertia about the center of mass is known; to find the moment of inertia about an axis parallel to an axis passing through the center of mass, use the Parallel Axis Theorem.
  • The theorem states that the distance from the new axis to the original axis passing through the center of mass is crucial.
  • Two parallel axes are needed, one passing through the center of mass and the other at a distance from it, to calculate the moment of inertia.
  • The theorem is applicable in both 2D and 3D bodies, with the Perpendicular Axis Theorem specific to 2D bodies.
  • For a rod with mass 'm' and length 'a', the moment of inertia about an axis passing through one end and perpendicular to its length is 'm*a^2/3'.
  • When four identical thin rods of mass 'm' and length 'l' form a square frame, the moment of inertia about an axis through the center of the square and perpendicular to its plane is calculated.
  • A thin rod of mass 'm' and length 'l' bent at its midpoint into two halves at a 90-degree angle requires determining the moment of inertia about an axis passing through the bending point and perpendicular to the plane defined by the two halves.
  • The moment of inertia of a uniform circular disk of mass 'm' and radius 'r' about an axis touching the disk at its diameter and normal to the disk is essential to calculate.
  • The moment of inertia for a uniform circular disk of radius 'r' and mass 'm' about an axis touching the disk at its diameter and normal to the disk is a key calculation.
  • Understanding the principles of moment of inertia and applying the Parallel Axis Theorem are crucial for accurate calculations in physics.

02:52:44

Moment of Inertia for Standard Bodies

  • The text discusses the Moment of Inception of a Uniform Circular Disk of mass m and radius r about an axis touching the disk at its diameter and normal to the disc and this axis.
  • The text emphasizes the removal of AXD from the required axis x x d.
  • The Center of Mass and the distance a a become equal to the radius r.
  • The Moment of Inertia of a Ring about an axis passing through the center of mass and perpendicular to the plane of the ring is m s / 2.
  • The value of a is r, resulting in m s so son 2 and 3/2 m.
  • The Theorem of Perpendicular axis is discussed, focusing on the perpendicular axis to a 2D object.
  • The text explains the Moment of Inertia About the axis for a ring, with the moment of inertia being m s / 2.
  • The Moment of Inertia of a Disk about an axis tangential to the circumference of the disk and parallel to its diameter is calculated to be 5m s / 4.
  • The text delves into the Moment of Inertia of Some Standard Bodies, including a rod, ring, and disk, providing formulas for each.
  • The Perpendicular Axis Theorem is derived for a rectangle, with the moment of inertia being m a sc b s up 6.

03:11:46

"Baba's Heart Alarm Rings for Ant"

  • Baba's life is in an ant, with his heart residing in the ant.
  • An alarm bell is connected to Baba's heart, ringing if any ant dies.
  • A random girl named Riya buries an ant, triggering the alarm bell.
  • The story is narrated through a song about an ant's death and Riya's involvement.
  • The concept of a square is discussed, focusing on axes and moments of inertia.
  • The moment of inertia for a hollow sphere and solid sphere is detailed.
  • The moment of inertia for a hollow cone and solid cone is explained.
  • The axis and moment of inertia for a hollow cylinder and solid cylinder are discussed.
  • The concept of a disc and its moment of inertia is elaborated upon.
  • The moment of inertia for a solid sphere about a tangent to the sphere is calculated.

03:30:27

Calculating r with 2m, solving 2mr=4m sk

  • The value of r is being calculated, with r being small and two times of m.
  • The equation 2mr = 4m sk is being solved to find the value of 4m sk.
  • Option D is confirmed as correct based on the calculations.
  • The moment of inertia about a triangular plate is discussed, focusing on the base and height.
  • The relationship between the mass of a triangle and a plate is explored to determine the moment of inertia.
  • The comparison between i1 and i2 is made based on the sizes of l1 and l2.
  • The concept of the radius of gyre is introduced, equating it to the moment of inertia.
  • The unit and dimension of the radius of gyre are explained.
  • The ratio of the radii of a circular disk and a circular ring with the same mass and radius is calculated.
  • The moment of inertia of the remaining portion of a disk after a circular hole is cut out is determined.

03:49:56

"Calculating Mass Density and Moment of Inertia"

  • Moment of inertia of the remaining part of the disc about a perpendicular axis passing through the center is calculated by subtracting the removed portion's moment of inertia from the total.
  • A circular hole with a diameter of r is cut through the disc, with the rim passing through the center, affecting the mass and radius of the disc.
  • The mass of the remaining part of the disc about a perpendicular axis passing through the center is determined by subtracting the moment of inertia of the removed portion from the total moment of inertia.
  • The mass density of the disc is calculated based on the surface mass density, considering the area and mass distribution across the disc.
  • The concept of mass density and its relation to surface mass density is explained, highlighting the proportional relationship between mass and radius squared.
  • Halving the radius of the disc results in a quarter of the original mass due to the proportional relationship between mass and radius squared.
  • The discussion on torque and moment of inertia concludes, with a schedule for upcoming physics classes provided for reference.
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