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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?
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A cord is wound round the circumference of a wheel of radius r. The ax...
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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)).
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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?
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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 2024 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 2024 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 2024 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 2024 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?.
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