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A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be
- a)[2mgh/I+mr2]1∕2
- b)[2mgh/I+2mr2]1∕2
- c)√2gh
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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A cord is wound round the circumference of wheel of radius r. The axis...
[2mgh/I mr2]1/2
We can use the principle of conservation of energy to solve this problem. Initially, the weight is at rest and has potential energy mgh. As it falls, it gains kinetic energy and the wheel gains rotational kinetic energy. At the bottom, all the potential energy is converted into kinetic energy of the weight and the rotational kinetic energy of the wheel. Therefore, we have:
mgh = (1/2)mv^2 + (1/2)Iω^2
where v is the velocity of the weight and ω is the angular velocity of the wheel.
We also know that v = rω, since the weight is attached to the circumference of the wheel.
Substituting this into the equation above and simplifying, we get:
mgh = (1/2)mv^2 + (1/2)(I/mr^2)v^2
Solving for v, we get:
v = [2mgh/(m + I/r^2)]
Substituting this back into v = rω, we get:
ω = [2mgh/(m + I/r^2)] / r
Simplifying, we get:
ω = [2mgh/I mr^2]1/2
Therefore, the answer is (a).
We can use the principle of conservation of energy to solve this problem. Initially, the weight is at rest and has potential energy mgh. As it falls, it gains kinetic energy and the wheel gains rotational kinetic energy. At the bottom, all the potential energy is converted into kinetic energy of the weight and the rotational kinetic energy of the wheel. Therefore, we have:
mgh = (1/2)mv^2 + (1/2)Iω^2
where v is the velocity of the weight and ω is the angular velocity of the wheel.
We also know that v = rω, since the weight is attached to the circumference of the wheel.
Substituting this into the equation above and simplifying, we get:
mgh = (1/2)mv^2 + (1/2)(I/mr^2)v^2
Solving for v, we get:
v = [2mgh/(m + I/r^2)]
Substituting this back into v = rω, we get:
ω = [2mgh/(m + I/r^2)] / r
Simplifying, we get:
ω = [2mgh/I mr^2]1/2
Therefore, the answer is (a).
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A cord is wound round the circumference of wheel of radius r. The axis...
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A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will bea)[2mgh/I+mr2]1∕2b)[2mgh/I+2mr2]1∕2c)√2ghd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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