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A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?
Correct answer is '25'. Can you explain this answer?
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A cylinder rotating at an angular speed of 50 rev/s is brought in cont...
fR = lα1 fR = lα2
α1 = α2 = 2p red/sec2
For A cylinder : ω = ω0 - αt ω = 100π - 2πt ...(i)
For B cylinder ω = ω0 αt ω0 = 0
ω = αt ω = 27πt ....(ii)
From (i) and (ii) ω= 100 π - ω
2ω = 100π
ω = 50π
From (ii) euqation 50 π = 2 πt
t = 25 sec
α1 = α2 = 2p red/sec2
For A cylinder : ω = ω0 - αt ω = 100π - 2πt ...(i)
For B cylinder ω = ω0 αt ω0 = 0
ω = αt ω = 27πt ....(ii)
From (i) and (ii) ω= 100 π - ω
2ω = 100π
ω = 50π
From (ii) euqation 50 π = 2 πt
t = 25 sec
Most Upvoted Answer
A cylinder rotating at an angular speed of 50 rev/s is brought in cont...
Solution:
Initial conditions:
- Angular speed of the rotating cylinder, ω1 = 50 rev/s
- Angular speed of the stationary cylinder, ω2 = 0 rev/s
- Common magnitude of acceleration and deceleration, α = 1 rev/s^2
We need to find the time taken for the two cylinders to have equal angular speed.
1. Find the angular acceleration of both cylinders
The torque acting on the rotating cylinder is given by:
τ = I α
where I is the moment of inertia and α is the angular acceleration.
The torque due to friction is:
τ = f r
where f is the force of friction and r is the radius of the cylinder.
Equating the two torques:
I α = f r
The moment of inertia of a solid cylinder is given by:
I = (1/2) m r^2
where m is the mass of the cylinder and r is the radius.
Substituting for I and simplifying:
α = (2/3) (f/m)
The angular acceleration of the rotating cylinder is therefore:
α1 = (2/3) (f/m)
The torque acting on the stationary cylinder is:
τ = I α
where I is the moment of inertia.
The moment of inertia of the stationary cylinder is the same as that of the rotating cylinder.
Therefore, the angular acceleration of the stationary cylinder is:
α2 = (2/3) (0/m) = 0
2. Find the time taken for the two cylinders to have equal angular speed
Let ω be the common angular speed of the two cylinders at any time t.
The angular speed of the rotating cylinder at time t is:
ω1 = ω + α t
The angular speed of the stationary cylinder at time t is:
ω2 = α t
Equating the two angular speeds and solving for t:
ω + α t = α t
ω = 0
Therefore, the time taken for the two cylinders to have equal angular speed is:
t = ω1/α = 50/1 = 25 seconds
Answer: The time taken for the two cylinders to have equal angular speed is 25 seconds.
Initial conditions:
- Angular speed of the rotating cylinder, ω1 = 50 rev/s
- Angular speed of the stationary cylinder, ω2 = 0 rev/s
- Common magnitude of acceleration and deceleration, α = 1 rev/s^2
We need to find the time taken for the two cylinders to have equal angular speed.
1. Find the angular acceleration of both cylinders
The torque acting on the rotating cylinder is given by:
τ = I α
where I is the moment of inertia and α is the angular acceleration.
The torque due to friction is:
τ = f r
where f is the force of friction and r is the radius of the cylinder.
Equating the two torques:
I α = f r
The moment of inertia of a solid cylinder is given by:
I = (1/2) m r^2
where m is the mass of the cylinder and r is the radius.
Substituting for I and simplifying:
α = (2/3) (f/m)
The angular acceleration of the rotating cylinder is therefore:
α1 = (2/3) (f/m)
The torque acting on the stationary cylinder is:
τ = I α
where I is the moment of inertia.
The moment of inertia of the stationary cylinder is the same as that of the rotating cylinder.
Therefore, the angular acceleration of the stationary cylinder is:
α2 = (2/3) (0/m) = 0
2. Find the time taken for the two cylinders to have equal angular speed
Let ω be the common angular speed of the two cylinders at any time t.
The angular speed of the rotating cylinder at time t is:
ω1 = ω + α t
The angular speed of the stationary cylinder at time t is:
ω2 = α t
Equating the two angular speeds and solving for t:
ω + α t = α t
ω = 0
Therefore, the time taken for the two cylinders to have equal angular speed is:
t = ω1/α = 50/1 = 25 seconds
Answer: The time taken for the two cylinders to have equal angular speed is 25 seconds.
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A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer?
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A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. 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 cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. 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 cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer?.
A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. 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 cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. 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 cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer?.
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A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer?, a detailed solution for A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer? has been provided alongside types of A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?Correct answer is '25'. Can you explain this answer? theory, EduRev gives you an
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