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Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is?
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Three mass particles A B and C having mass m 2m and 3m respectively ar...
Kinetic Energy of the System
Given:
Mass of particle A = m
Mass of particle B = 2m
Mass of particle C = 3m
Mass of the ring = m
Radius of the ring = r
Velocity of the center of the ring = v'
To find the kinetic energy of the system, we need to consider the kinetic energy of each individual particle and the kinetic energy of the ring.
Kinetic Energy of Particle A:
The kinetic energy of a particle is given by the formula KE = (1/2)mv^2, where m is the mass of the particle and v is its velocity.
The mass of particle A is m, and its velocity is the same as the velocity of the center of the ring, v'. Therefore, the kinetic energy of particle A is (1/2)m(v')^2.
Kinetic Energy of Particle B:
Similarly, the mass of particle B is 2m, and its velocity is also v'. Therefore, the kinetic energy of particle B is (1/2)(2m)(v')^2 = 2m(v')^2.
Kinetic Energy of Particle C:
The mass of particle C is 3m, and its velocity is also v'. Therefore, the kinetic energy of particle C is (1/2)(3m)(v')^2 = 3m(v')^2.
Kinetic Energy of the Ring:
The kinetic energy of a rotating object is given by the formula KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
For a ring rolling without slipping, the moment of inertia is given by I = MR^2, where M is the mass of the ring and R is its radius.
The mass of the ring is m and its radius is r, so the moment of inertia of the ring is I = m(r^2).
The angular velocity of the ring can be calculated using the relation v = ωR, where v is the velocity of the center of the ring and R is its radius.
Therefore, ω = v/r.
Substituting the values of I and ω into the formula for kinetic energy, we get KE = (1/2)m(r^2)(v/r)^2 = (1/2)mv^2.
Total Kinetic Energy of the System:
The total kinetic energy of the system is the sum of the kinetic energies of all the particles and the ring.
KE_total = KE_A + KE_B + KE_C + KE_ring
= (1/2)m(v')^2 + 2m(v')^2 + 3m(v')^2 + (1/2)mv^2
= (1/2)m(v')^2 + 6m(v')^2 + (1/2)mv^2
= (7/2)m(v')^2 + (1/2)mv^2
Therefore, the kinetic energy of the system is given by (7/2)m(v')^2 + (1/2)mv^2.
Given:
Mass of particle A = m
Mass of particle B = 2m
Mass of particle C = 3m
Mass of the ring = m
Radius of the ring = r
Velocity of the center of the ring = v'
To find the kinetic energy of the system, we need to consider the kinetic energy of each individual particle and the kinetic energy of the ring.
Kinetic Energy of Particle A:
The kinetic energy of a particle is given by the formula KE = (1/2)mv^2, where m is the mass of the particle and v is its velocity.
The mass of particle A is m, and its velocity is the same as the velocity of the center of the ring, v'. Therefore, the kinetic energy of particle A is (1/2)m(v')^2.
Kinetic Energy of Particle B:
Similarly, the mass of particle B is 2m, and its velocity is also v'. Therefore, the kinetic energy of particle B is (1/2)(2m)(v')^2 = 2m(v')^2.
Kinetic Energy of Particle C:
The mass of particle C is 3m, and its velocity is also v'. Therefore, the kinetic energy of particle C is (1/2)(3m)(v')^2 = 3m(v')^2.
Kinetic Energy of the Ring:
The kinetic energy of a rotating object is given by the formula KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
For a ring rolling without slipping, the moment of inertia is given by I = MR^2, where M is the mass of the ring and R is its radius.
The mass of the ring is m and its radius is r, so the moment of inertia of the ring is I = m(r^2).
The angular velocity of the ring can be calculated using the relation v = ωR, where v is the velocity of the center of the ring and R is its radius.
Therefore, ω = v/r.
Substituting the values of I and ω into the formula for kinetic energy, we get KE = (1/2)m(r^2)(v/r)^2 = (1/2)mv^2.
Total Kinetic Energy of the System:
The total kinetic energy of the system is the sum of the kinetic energies of all the particles and the ring.
KE_total = KE_A + KE_B + KE_C + KE_ring
= (1/2)m(v')^2 + 2m(v')^2 + 3m(v')^2 + (1/2)mv^2
= (1/2)m(v')^2 + 6m(v')^2 + (1/2)mv^2
= (7/2)m(v')^2 + (1/2)mv^2
Therefore, the kinetic energy of the system is given by (7/2)m(v')^2 + (1/2)mv^2.
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Three mass particles A B and C having mass m 2m and 3m respectively ar...
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Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is?
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Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is?.
Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three mass particles A B and C having mass m 2m and 3m respectively are rigidly attached to a ring of mass m and radius r which rolls on a horizontal surface without slipping . At a certain instant the velocity of the centre of the ring is v'.the kinetic energy of the system is?.
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