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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1
about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,
are gently placed symmetrically on the disc in such a manner that they are touching each other along the
axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest
relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) of
the system is
about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,
are gently placed symmetrically on the disc in such a manner that they are touching each other along the
axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest
relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) of
the system is
Correct answer is '8'. Can you explain this answer?
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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating wit...
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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating wit...
Given:
Mass of the disc (m1) = 50 kg
Radius of the disc (r1) = 0.4 m
Angular velocity of the disc (ω1) = 10 rad/s
Mass of each ring (m2) = 6.25 kg
Radius of each ring (r2) = 0.2 m
Approach:
To find the new angular velocity of the system, we can use the principle of conservation of angular momentum. According to this principle, the total angular momentum of a system remains conserved if no external torque acts on it.
The initial angular momentum of the system is given by:
L_initial = I1 * ω1 + 2 * I2 * 0, where I1 is the moment of inertia of the disc and I2 is the moment of inertia of each ring.
The final angular momentum of the system is given by:
L_final = I_final * ω_final, where I_final is the moment of inertia of the system (disc + rings) and ω_final is the final angular velocity of the system.
Since the rings are symmetrically placed on the disc along its axis, the moment of inertia of the system can be calculated as:
I_final = I1 + 2 * I2, where I1 is the moment of inertia of the disc and I2 is the moment of inertia of each ring.
Calculation:
The moment of inertia of a disc about its axis of rotation is given by:
I1 = (1/2) * m1 * r1^2
Substituting the given values:
I1 = (1/2) * 50 * 0.4^2 = 2 kg.m^2
The moment of inertia of a ring about its axis of rotation is given by:
I2 = m2 * r2^2
Substituting the given values:
I2 = 6.25 * 0.2^2 = 0.25 kg.m^2
The moment of inertia of the system is:
I_final = I1 + 2 * I2 = 2 + 2 * 0.25 = 2.5 kg.m^2
Using the conservation of angular momentum, we can equate the initial and final angular momenta:
I1 * ω1 + 2 * I2 * 0 = I_final * ω_final
Substituting the values:
2 * 10 + 2 * 0.25 * 0 = 2.5 * ω_final
Simplifying the equation:
20 = 2.5 * ω_final
Solving for ω_final:
ω_final = 20 / 2.5 = 8 rad/s
Therefore, the new angular velocity of the system is 8 rad/s.
Mass of the disc (m1) = 50 kg
Radius of the disc (r1) = 0.4 m
Angular velocity of the disc (ω1) = 10 rad/s
Mass of each ring (m2) = 6.25 kg
Radius of each ring (r2) = 0.2 m
Approach:
To find the new angular velocity of the system, we can use the principle of conservation of angular momentum. According to this principle, the total angular momentum of a system remains conserved if no external torque acts on it.
The initial angular momentum of the system is given by:
L_initial = I1 * ω1 + 2 * I2 * 0, where I1 is the moment of inertia of the disc and I2 is the moment of inertia of each ring.
The final angular momentum of the system is given by:
L_final = I_final * ω_final, where I_final is the moment of inertia of the system (disc + rings) and ω_final is the final angular velocity of the system.
Since the rings are symmetrically placed on the disc along its axis, the moment of inertia of the system can be calculated as:
I_final = I1 + 2 * I2, where I1 is the moment of inertia of the disc and I2 is the moment of inertia of each ring.
Calculation:
The moment of inertia of a disc about its axis of rotation is given by:
I1 = (1/2) * m1 * r1^2
Substituting the given values:
I1 = (1/2) * 50 * 0.4^2 = 2 kg.m^2
The moment of inertia of a ring about its axis of rotation is given by:
I2 = m2 * r2^2
Substituting the given values:
I2 = 6.25 * 0.2^2 = 0.25 kg.m^2
The moment of inertia of the system is:
I_final = I1 + 2 * I2 = 2 + 2 * 0.25 = 2.5 kg.m^2
Using the conservation of angular momentum, we can equate the initial and final angular momenta:
I1 * ω1 + 2 * I2 * 0 = I_final * ω_final
Substituting the values:
2 * 10 + 2 * 0.25 * 0 = 2.5 * ω_final
Simplifying the equation:
20 = 2.5 * ω_final
Solving for ω_final:
ω_final = 20 / 2.5 = 8 rad/s
Therefore, the new angular velocity of the system is 8 rad/s.
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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer?
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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer?.
A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer?.
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A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer?, a detailed solution for A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? has been provided alongside types of A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) ofthe system isCorrect answer is '8'. Can you explain this answer? tests, examples and also practice JEE tests.
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