A mass is tied to a thin, light string that is wound around a cylinder. The mass is released. There is a gravitational force pulling it downward, and the tension in the string is pulling it upward. This difference between the gravitational force and the tension can be calculated to find the net forcRead more
A mass is tied to a thin, light string that is wound around a cylinder. The mass is released. There is a gravitational force pulling it downward, and the tension in the string is pulling it upward. This difference between the gravitational force and the tension can be calculated to find the net force on the mass. By Newton’s second law, this net force must equal the product of the mass and its linear acceleration.
There’s also tension in the string creating a torque in the cylinder which can be represented in terms of tension and radius of the cylinder. The angular acceleration is found to be associated with the linear acceleration of mass by the radius of the cylinder.
We can obtain the linear acceleration of the falling mass and the tension in the string by considering these relationships. The acceleration can be written as a fraction of the acceleration due to gravity, showing that it is smaller than the acceleration due to gravity. This happens because part of the gravitational force is being utilized to generate rotational motion in the cylinder, meaning that there will be a smaller linear acceleration of the mass than if it were falling freely. This whole system demonstrates principles of rotational dynamics and linear motion.
In finding the linear acceleration of a rolling cylinder on an inclined plane, we have the forces involved-the weight of the cylinder and the frictional force. The weight can be divided into two: one along the incline and the other perpendicular to it. The friction force opposes the motion to avoidRead more
In finding the linear acceleration of a rolling cylinder on an inclined plane, we have the forces involved-the weight of the cylinder and the frictional force. The weight can be divided into two: one along the incline and the other perpendicular to it. The friction force opposes the motion to avoid sliding.
Using Newton’s second law, we analyze the net force acting along the incline on the cylinder. The net force is the difference between the component of gravitational force pulling it down the slope and the frictional force acting up the slope. This net force is also related to the linear acceleration of the cylinder.
Next, we analyze the cylinder’s rotational motion where the frictional force causes a torque around the center of mass. Torque affects the cylinder’s angular acceleration. Linear acceleration and angular acceleration for rolling without slipping are in fact related.
Integrating these considerations shows that the linear acceleration of the rolling cylinder down the incline is a fraction of gravitational acceleration. The no slipping requirement for the rolling cylinder further depends on the coefficient of static friction and the angle of the incline so that the frictional force is always quite sufficient to prevent slipping.
When a uniform square plate rotates about an axis in its plane, the moment of inertia depends on the orientation of the axis relative to the sides of the plate. Given that the moment of inertia about an axis parallel to two sides is denoted as l, and the axis CD makes an angle θ with this axis, theRead more
When a uniform square plate rotates about an axis in its plane, the moment of inertia depends on the orientation of the axis relative to the sides of the plate. Given that the moment of inertia about an axis parallel to two sides is denoted as l, and the axis CD makes an angle θ with this axis, the moment of inertia about axis CD is l/cos² θ.
This is because the rotational inertia depends on the distribution of mass relative to the axis of rotation. The cosine squared term accounts for the projection of the mass distribution onto the new axis.
Thus, the correct answer is that the moment of inertia about axis CD is l /cos² θ.
If the ball were to strike the floor, hit the ground and rebound as in an inelastic collision, it would follow that the total momentum of the Earth and the ball is conserved. However, in the inelastic collision, some of its kinetic energy becomes other types of energy like heat or sound that causesRead more
If the ball were to strike the floor, hit the ground and rebound as in an inelastic collision, it would follow that the total momentum of the Earth and the ball is conserved. However, in the inelastic collision, some of its kinetic energy becomes other types of energy like heat or sound that causes loss of mechanical energy but the system with both Earth and the ball together conserves their total momentum.
This is a basic principle in physics, which states that the total momentum of a closed system remains constant if no external forces act upon it. In this case, the ball and the Earth constitute a closed system, and the internal forces during the collision do not affect the total momentum.
Note that, although momentum is conserved, mechanical energy is not in inelastic collisions because kinetic energy is converted into other forms of energy. Thus, the correct answer is that the total momentum of the ball and the Earth is conserved.
The angular momentum of a particle is fundamentally linked to its linear momentum and the concept of the moment arm, which is the perpendicular distance from the line of action of the linear momentum to the axis of rotation. Angular momentum can be defined as the product of the particle's linear momRead more
The angular momentum of a particle is fundamentally linked to its linear momentum and the concept of the moment arm, which is the perpendicular distance from the line of action of the linear momentum to the axis of rotation. Angular momentum can be defined as the product of the particle’s linear momentum and this moment arm. This relationship points out how the rotational effect of a particle is dependent, not only on its velocity but also on its distance from the axis about which it rotates.
In addition, it is worthwhile to note that angular momentum results purely from the angular component of linear momentum. The linear momentum of a particle can be divided into two components. One component is along the radius, known as the radial component, and the other is perpendicular to the radius, known as the tangential component. Since the radial component acts along the radius, it does not contribute to angular momentum because it does not produce any rotational effect. The tangential component of the linear momentum would be the one that impacts the angular momentum as it creates rotation around the axis.
Angular momentum is thus the product of the linear momentum and the moment arm, and only the tangential component of linear momentum produces the motion in an angular direction.
A light string is wound round a cylinder and carries a mass tied to it at the free end. When the mass is released, calculate (a) the linear acceleration of the descending mass and (b) the angular acceleration of the cylinder and (c) the tension in the string. Show that the acceleration of mass is less than g.
A mass is tied to a thin, light string that is wound around a cylinder. The mass is released. There is a gravitational force pulling it downward, and the tension in the string is pulling it upward. This difference between the gravitational force and the tension can be calculated to find the net forcRead more
A mass is tied to a thin, light string that is wound around a cylinder. The mass is released. There is a gravitational force pulling it downward, and the tension in the string is pulling it upward. This difference between the gravitational force and the tension can be calculated to find the net force on the mass. By Newton’s second law, this net force must equal the product of the mass and its linear acceleration.
There’s also tension in the string creating a torque in the cylinder which can be represented in terms of tension and radius of the cylinder. The angular acceleration is found to be associated with the linear acceleration of mass by the radius of the cylinder.
We can obtain the linear acceleration of the falling mass and the tension in the string by considering these relationships. The acceleration can be written as a fraction of the acceleration due to gravity, showing that it is smaller than the acceleration due to gravity. This happens because part of the gravitational force is being utilized to generate rotational motion in the cylinder, meaning that there will be a smaller linear acceleration of the mass than if it were falling freely. This whole system demonstrates principles of rotational dynamics and linear motion.
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See lessObtain the expression for the linear acceleration of a cylinder rolling down inclined place and hence find the condition for the cylinder to roll down without slipping.
In finding the linear acceleration of a rolling cylinder on an inclined plane, we have the forces involved-the weight of the cylinder and the frictional force. The weight can be divided into two: one along the incline and the other perpendicular to it. The friction force opposes the motion to avoidRead more
In finding the linear acceleration of a rolling cylinder on an inclined plane, we have the forces involved-the weight of the cylinder and the frictional force. The weight can be divided into two: one along the incline and the other perpendicular to it. The friction force opposes the motion to avoid sliding.
Using Newton’s second law, we analyze the net force acting along the incline on the cylinder. The net force is the difference between the component of gravitational force pulling it down the slope and the frictional force acting up the slope. This net force is also related to the linear acceleration of the cylinder.
Next, we analyze the cylinder’s rotational motion where the frictional force causes a torque around the center of mass. Torque affects the cylinder’s angular acceleration. Linear acceleration and angular acceleration for rolling without slipping are in fact related.
Integrating these considerations shows that the linear acceleration of the rolling cylinder down the incline is a fraction of gravitational acceleration. The no slipping requirement for the rolling cylinder further depends on the coefficient of static friction and the angle of the incline so that the frictional force is always quite sufficient to prevent slipping.
see more :- https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessLet l be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to
When a uniform square plate rotates about an axis in its plane, the moment of inertia depends on the orientation of the axis relative to the sides of the plate. Given that the moment of inertia about an axis parallel to two sides is denoted as l, and the axis CD makes an angle θ with this axis, theRead more
When a uniform square plate rotates about an axis in its plane, the moment of inertia depends on the orientation of the axis relative to the sides of the plate. Given that the moment of inertia about an axis parallel to two sides is denoted as l, and the axis CD makes an angle θ with this axis, the moment of inertia about axis CD is l/cos² θ.
This is because the rotational inertia depends on the distribution of mass relative to the axis of rotation. The cosine squared term accounts for the projection of the mass distribution onto the new axis.
Thus, the correct answer is that the moment of inertia about axis CD is l /cos² θ.
Checkout here : – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessA ball hits the floor and rebounds after an inelastic collision. In this case
If the ball were to strike the floor, hit the ground and rebound as in an inelastic collision, it would follow that the total momentum of the Earth and the ball is conserved. However, in the inelastic collision, some of its kinetic energy becomes other types of energy like heat or sound that causesRead more
If the ball were to strike the floor, hit the ground and rebound as in an inelastic collision, it would follow that the total momentum of the Earth and the ball is conserved. However, in the inelastic collision, some of its kinetic energy becomes other types of energy like heat or sound that causes loss of mechanical energy but the system with both Earth and the ball together conserves their total momentum.
This is a basic principle in physics, which states that the total momentum of a closed system remains constant if no external forces act upon it. In this case, the ball and the Earth constitute a closed system, and the internal forces during the collision do not affect the total momentum.
Note that, although momentum is conserved, mechanical energy is not in inelastic collisions because kinetic energy is converted into other forms of energy. Thus, the correct answer is that the total momentum of the ball and the Earth is conserved.
See more : – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessShow that the angular momentum of a particle is the product of its linear momentum and the moment arm. Also show that the angular momentum is produced only by the angular component of linear momentum.
The angular momentum of a particle is fundamentally linked to its linear momentum and the concept of the moment arm, which is the perpendicular distance from the line of action of the linear momentum to the axis of rotation. Angular momentum can be defined as the product of the particle's linear momRead more
The angular momentum of a particle is fundamentally linked to its linear momentum and the concept of the moment arm, which is the perpendicular distance from the line of action of the linear momentum to the axis of rotation. Angular momentum can be defined as the product of the particle’s linear momentum and this moment arm. This relationship points out how the rotational effect of a particle is dependent, not only on its velocity but also on its distance from the axis about which it rotates.
In addition, it is worthwhile to note that angular momentum results purely from the angular component of linear momentum. The linear momentum of a particle can be divided into two components. One component is along the radius, known as the radial component, and the other is perpendicular to the radius, known as the tangential component. Since the radial component acts along the radius, it does not contribute to angular momentum because it does not produce any rotational effect. The tangential component of the linear momentum would be the one that impacts the angular momentum as it creates rotation around the axis.
Angular momentum is thus the product of the linear momentum and the moment arm, and only the tangential component of linear momentum produces the motion in an angular direction.
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See less