When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass ofRead more
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass of the system.
Here the rod is of length 1.4 m. The point masses are 0.3 kg and 0.7 kg located at the opposite ends of this rod. It is found that the center of mass is placed at a distance of about 0.98 m from the 0.3 kg mass. So to minimize the work done in the rotation, this axis of rotation should pass through this point.
Positioning the axis at the center of mass minimizes the rotational inertia of the system, making the rotation more efficient. This is one of the approaches in physics to make calculations easier and reduce the energy required to start rotational motion.
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocityRead more
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocity of 14 m/s towards the minor block, given that the little block is assumed to be moving on a horizontal frictionless ground.
It is the average position of the entire mass in the system weighted by their respective velocities. The velocity of the center of mass for this system would depend on the contributions from each block’s mass and its respective velocity. In this case, because the light block has no movement at all it contributes nothing to the velocity of the center of mass. Its movement is dominated by that of the heavier block since the former has more mass and higher speed.
By adding both blocks’ masses and velocities together, the center of mass’s velocity is found to be 10 m/s. This represents the motion of the system in a frictionless surface. The center of mass’s velocity is only changed when acted on by an outside force according to the principles of physics.
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with aRead more
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with a mass of m/4 and the other with a mass of 3m/4.
Given after the explosion that the smaller fragment is located at y = +15 cm, use the idea that the overall center of mass for the system must remain at the same vertical position as before the explosion, which was at y = 0.
The position of the center of mass is determined by the masses and their respective positions. Since the smaller fragment is at y = +15 cm, the position of the larger fragment must be such that it balances this position according to their mass ratio. By conservation, we determine that the larger fragment must be positioned at y = -5 cm. This negative position denotes that it lies below the initial center of mass level, making sure the global system maintains the center of mass position after detonation.
In uniform circular motion, the particle moves along a circular path at constant speed. For such a particle, angular momentum is conserved about any point in the plane of the circle if no external torques act on the system; thus, the angular momentum is constant about any point in the plane of the cRead more
In uniform circular motion, the particle moves along a circular path at constant speed. For such a particle, angular momentum is conserved about any point in the plane of the circle if no external torques act on the system; thus, the angular momentum is constant about any point in the plane of the circle.
The conservation of angular momentum is one of the principles in physics. According to it, if there is no external torque acting on the system, its angular momentum will be conserved. For a uniform circular motion, the motion of the particle is restricted to a plane and the forces that act upon it are internal to the system. Thus, its angular momentum will be conserved about any point within that plane.
It should be noted that even though the angular momentum is conserved about points that are inside the plane of the circle, the value of the angular momentum is dependent on the selection of the reference point. On the other hand, the reason that it remains conserved about any point in the plane results directly from the lack of external torques.
When a particle moves in a circular path with decreasing linear speed, it experiences a tangential acceleration directed opposite to its velocity, causing it to slow down. This deceleration results in a reduction of the particle's angular momentum over time. Thus, the angular momentum of the particlRead more
When a particle moves in a circular path with decreasing linear speed, it experiences a tangential acceleration directed opposite to its velocity, causing it to slow down. This deceleration results in a reduction of the particle’s angular momentum over time. Thus, the angular momentum of the particle is not conserved in this scenario.
If the speed of the particle were constant, then it would only have centripetal acceleration toward the center of the circle, so it would have a constant angular momentum. In this case, the existence of tangential acceleration means that the speed of the particle, and hence its angular momentum, will change.
Therefore, the correct statement is that the angular momentum of the particle is not conserved while it moves in a circular path with decreasing linear speed.
Two particles, A and B, start from rest, moving toward each other under mutual forces of attraction. In this case, the center of mass of the system continues to have zero speed. This is so because the system of two particles is isolated; no external forces are acting on it. In such systems, the motiRead more
Two particles, A and B, start from rest, moving toward each other under mutual forces of attraction. In this case, the center of mass of the system continues to have zero speed. This is so because the system of two particles is isolated; no external forces are acting on it. In such systems, the motion of the center of mass is determined only by external forces. Since there are none in this case, the center of mass remains stationary or maintains a constant velocity. Initially, both particles are at rest, so the velocity of the center of mass is zero.
As they move toward each other because of mutual attraction, their velocities change, but these changes are such that the internal forces do not affect the center of mass. Internal forces always act in equal and opposite directions, thus canceling out their effects on the system as a whole. Hence, even in the case of particle A travelling at speed v and particle B traveling at twice that speed (2v) the center of mass of this system does not gain any velocity to move. Such a scenario is in fact an exercise in the fundamental conservation of momentum concept in isolated systems.
(i) Investing in Improved Healthcare Infrastructure: - Enhances access to medical facilities, preventive care, and essential treatments. - Enables early diagnosis, effective disease management, and prevention. - Reduces the burden of preventable diseases and chronic conditions. - Lowers mortality raRead more
(i) Investing in Improved Healthcare Infrastructure:
– Enhances access to medical facilities, preventive care, and essential treatments.
– Enables early diagnosis, effective disease management, and prevention.
– Reduces the burden of preventable diseases and chronic conditions.
– Lowers mortality rates and improves overall public health.
(ii) Relationship between Population Trends and Poverty Reduction:
– Balanced population growth and age structure spur economic growth and development.
– Proper rural-urban distribution offers more opportunities and resources, reducing poverty.
– High population growth, skewed demographics, and rural-urban disparities strain resources, hindering poverty alleviation efforts.
(iii) Mutual Relationship between Poverty and Population Dynamics:
– Population dynamics (growth, age structure, distribution) influence poverty levels.
– Poverty impacts demographic trends, affecting birth rates, migration, and access to services.
– Interplay between poverty and population dynamics shapes development outcomes, highlighting their mutual influence on each other.
Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass ofRead more
When a rod having point masses affixed to both ends rotates, the work which must be performed to set its rotation going, and also sustain it, will depend on how the mass distribution is relative to the axis of rotation. So, in order to minimize such work, it should pass through the center of mass of the system.
Here the rod is of length 1.4 m. The point masses are 0.3 kg and 0.7 kg located at the opposite ends of this rod. It is found that the center of mass is placed at a distance of about 0.98 m from the 0.3 kg mass. So to minimize the work done in the rotation, this axis of rotation should pass through this point.
Positioning the axis at the center of mass minimizes the rotational inertia of the system, making the rotation more efficient. This is one of the approaches in physics to make calculations easier and reduce the energy required to start rotational motion.
For more click here : – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessTwo blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocityRead more
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocity of 14 m/s towards the minor block, given that the little block is assumed to be moving on a horizontal frictionless ground.
It is the average position of the entire mass in the system weighted by their respective velocities. The velocity of the center of mass for this system would depend on the contributions from each block’s mass and its respective velocity. In this case, because the light block has no movement at all it contributes nothing to the velocity of the center of mass. Its movement is dominated by that of the heavier block since the former has more mass and higher speed.
By adding both blocks’ masses and velocities together, the center of mass’s velocity is found to be 10 m/s. This represents the motion of the system in a frictionless surface. The center of mass’s velocity is only changed when acted on by an outside force according to the principles of physics.
Checkout this for more information: – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessAn isolated particle of mass m is moving in horizontal in horizontal plane (x-y) along the x-axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with aRead more
The isolated particle of mass m is moving horizontally along the x-axis and suddenly explodes into two fragments. The center of mass, according to the principle of conservation of momentum, remains unchanged if no external forces act on the system. The particle breaks into two fragments, one with a mass of m/4 and the other with a mass of 3m/4.
Given after the explosion that the smaller fragment is located at y = +15 cm, use the idea that the overall center of mass for the system must remain at the same vertical position as before the explosion, which was at y = 0.
The position of the center of mass is determined by the masses and their respective positions. Since the smaller fragment is at y = +15 cm, the position of the larger fragment must be such that it balances this position according to their mass ratio. By conservation, we determine that the larger fragment must be positioned at y = -5 cm. This negative position denotes that it lies below the initial center of mass level, making sure the global system maintains the center of mass position after detonation.
Click here for more: – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessA particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?
In uniform circular motion, the particle moves along a circular path at constant speed. For such a particle, angular momentum is conserved about any point in the plane of the circle if no external torques act on the system; thus, the angular momentum is constant about any point in the plane of the cRead more
In uniform circular motion, the particle moves along a circular path at constant speed. For such a particle, angular momentum is conserved about any point in the plane of the circle if no external torques act on the system; thus, the angular momentum is constant about any point in the plane of the circle.
The conservation of angular momentum is one of the principles in physics. According to it, if there is no external torque acting on the system, its angular momentum will be conserved. For a uniform circular motion, the motion of the particle is restricted to a plane and the forces that act upon it are internal to the system. Thus, its angular momentum will be conserved about any point within that plane.
It should be noted that even though the angular momentum is conserved about points that are inside the plane of the circle, the value of the angular momentum is dependent on the selection of the reference point. On the other hand, the reason that it remains conserved about any point in the plane results directly from the lack of external torques.
Check here for more details: – https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessA particle is confined to rotate in a circular path with decreasing liner speed. Then which of the following is correct?
When a particle moves in a circular path with decreasing linear speed, it experiences a tangential acceleration directed opposite to its velocity, causing it to slow down. This deceleration results in a reduction of the particle's angular momentum over time. Thus, the angular momentum of the particlRead more
When a particle moves in a circular path with decreasing linear speed, it experiences a tangential acceleration directed opposite to its velocity, causing it to slow down. This deceleration results in a reduction of the particle’s angular momentum over time. Thus, the angular momentum of the particle is not conserved in this scenario.
If the speed of the particle were constant, then it would only have centripetal acceleration toward the center of the circle, so it would have a constant angular momentum. In this case, the existence of tangential acceleration means that the speed of the particle, and hence its angular momentum, will change.
Therefore, the correct statement is that the angular momentum of the particle is not conserved while it moves in a circular path with decreasing linear speed.
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See lessTwo particles A and B, initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2v, the speed of the centre of mass of the system is
Two particles, A and B, start from rest, moving toward each other under mutual forces of attraction. In this case, the center of mass of the system continues to have zero speed. This is so because the system of two particles is isolated; no external forces are acting on it. In such systems, the motiRead more
Two particles, A and B, start from rest, moving toward each other under mutual forces of attraction. In this case, the center of mass of the system continues to have zero speed. This is so because the system of two particles is isolated; no external forces are acting on it. In such systems, the motion of the center of mass is determined only by external forces. Since there are none in this case, the center of mass remains stationary or maintains a constant velocity. Initially, both particles are at rest, so the velocity of the center of mass is zero.
As they move toward each other because of mutual attraction, their velocities change, but these changes are such that the internal forces do not affect the center of mass. Internal forces always act in equal and opposite directions, thus canceling out their effects on the system as a whole. Hence, even in the case of particle A travelling at speed v and particle B traveling at twice that speed (2v) the center of mass of this system does not gain any velocity to move. Such a scenario is in fact an exercise in the fundamental conservation of momentum concept in isolated systems.
Click here for more:- https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/
See lessRead the passage given below and answer the questions followed:
(i) Investing in Improved Healthcare Infrastructure: - Enhances access to medical facilities, preventive care, and essential treatments. - Enables early diagnosis, effective disease management, and prevention. - Reduces the burden of preventable diseases and chronic conditions. - Lowers mortality raRead more
(i) Investing in Improved Healthcare Infrastructure:
– Enhances access to medical facilities, preventive care, and essential treatments.
– Enables early diagnosis, effective disease management, and prevention.
– Reduces the burden of preventable diseases and chronic conditions.
– Lowers mortality rates and improves overall public health.
(ii) Relationship between Population Trends and Poverty Reduction:
– Balanced population growth and age structure spur economic growth and development.
– Proper rural-urban distribution offers more opportunities and resources, reducing poverty.
– High population growth, skewed demographics, and rural-urban disparities strain resources, hindering poverty alleviation efforts.
(iii) Mutual Relationship between Poverty and Population Dynamics:
See less– Population dynamics (growth, age structure, distribution) influence poverty levels.
– Poverty impacts demographic trends, affecting birth rates, migration, and access to services.
– Interplay between poverty and population dynamics shapes development outcomes, highlighting their mutual influence on each other.