NEET Exam > NEET Questions > A cord is wound round the circumference of a ...
Start Learning for Free
A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will be
- a)√2gh/I+mr
- b)√2mgh/I+mr2
- c)√2mgh/I+Mr2
- d)√2gh
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A cord is wound round the circumference of a wheel of radius r. The ax...
Most Upvoted Answer
A cord is wound round the circumference of a wheel of radius r. The ax...
To find the angular velocity of the wheel after the weight falls through a distance h, we need to consider the conservation of mechanical energy.
When the weight falls through a distance h, it loses potential energy equal to mgh. This energy is converted into kinetic energy of the weight and rotational kinetic energy of the wheel.
The kinetic energy of the weight is given by (1/2)mv^2, where v is the linear velocity of the weight. Since the weight falls from rest, the initial linear velocity is 0.
The rotational kinetic energy of the wheel is given by (1/2)Iω^2, where ω is the angular velocity of the wheel.
Since there is no slipping between the cord and the wheel, the linear velocity of the weight is equal to the tangential velocity of a point on the wheel's circumference, which is given by v = ωr.
Using these relations, we can write the conservation of mechanical energy equation as:
mgh = (1/2)mv^2 + (1/2)Iω^2
Substituting v = ωr, we get:
mgh = (1/2)m(ωr)^2 + (1/2)Iω^2
Simplifying, we have:
mgh = (1/2)mω^2r^2 + (1/2)Iω^2
Rearranging, we get:
mgh = (1/2)(mω^2r^2 + Iω^2)
Factoring out ω^2, we have:
mgh = (1/2)ω^2(mr^2 + I)
Solving for ω^2, we get:
ω^2 = (2mgh)/(mr^2 + I)
Taking the square root of both sides, we have:
ω = √((2mgh)/(mr^2 + I))
Therefore, the angular velocity of the wheel after the weight falls through a distance h is √((2mgh)/(mr^2 + I)).
When the weight falls through a distance h, it loses potential energy equal to mgh. This energy is converted into kinetic energy of the weight and rotational kinetic energy of the wheel.
The kinetic energy of the weight is given by (1/2)mv^2, where v is the linear velocity of the weight. Since the weight falls from rest, the initial linear velocity is 0.
The rotational kinetic energy of the wheel is given by (1/2)Iω^2, where ω is the angular velocity of the wheel.
Since there is no slipping between the cord and the wheel, the linear velocity of the weight is equal to the tangential velocity of a point on the wheel's circumference, which is given by v = ωr.
Using these relations, we can write the conservation of mechanical energy equation as:
mgh = (1/2)mv^2 + (1/2)Iω^2
Substituting v = ωr, we get:
mgh = (1/2)m(ωr)^2 + (1/2)Iω^2
Simplifying, we have:
mgh = (1/2)mω^2r^2 + (1/2)Iω^2
Rearranging, we get:
mgh = (1/2)(mω^2r^2 + Iω^2)
Factoring out ω^2, we have:
mgh = (1/2)ω^2(mr^2 + I)
Solving for ω^2, we get:
ω^2 = (2mgh)/(mr^2 + I)
Taking the square root of both sides, we have:
ω = √((2mgh)/(mr^2 + I))
Therefore, the angular velocity of the wheel after the weight falls through a distance h is √((2mgh)/(mr^2 + I)).
|
Explore Courses for NEET exam
|
|
Top Courses for NEETView all
Question Description
A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer?.
A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)√2gh/I+mrb)√2mgh/I+mr2c)√2mgh/I+Mr2d)√2ghCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup