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A cord is wound around the circumference...

A cord is wound around the circumference of wheel of radius r. The axis of the wheel is horizontal and `MI` 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

`sqrt((2gh)/(I+mr^(2)))`

B

`((2mgh)/(I+mr^(2)))^(1//2)`

C

`((2mgh)/(I+2mr^(2)))^(1//2)`

D

`sqrt(2gh)`

Text Solution

Verified by Experts

The correct Answer is:
B

Velocity, `v = r omega`
Decrease in gravitational potential energy = Increase in kinetic energy

`:. mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)=(1)/(2)Iomega^(2)+(1)/(2)m(r omega)^(2)`
`:.` Angular velocity, `omega=sqrt((2mgh)/(I+mr^(2)))`.
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Knowledge Check

  • 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 l. 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
    `sqrt((2gh)/(I + mr))`
    B
    `sqrt((2mgh)/(I + mr^(2)))`
    C
    `sqrt((2mgh)/(I+ 2mr^(2)))`
    D
    `sqrt(2gh)`
  • A cord is wound around the circumference of wheel of radius r the axis of the wheel is horizontal and moment of inerita about it is / The weight mg is attached to the end of the cord and falls from rest after falling throgh a distance h the angular velocity of the wheel will be

    A
    `[mgh]^(1//2)`
    B
    `[(2mgh)/(I+2mr^(2))]^(1//2)`
    C
    `[(2mgh)/(I+mr^(2))]^(1//2)`
    D
    `[(mgh)/(I+2mr^(2))]^(1//2)`
  • A cord is wound round the circumference of wheel of radius r . The axis of the wheel is horizontal and fixed 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
    `sqrt((2 gh)/(I + mr))`
    B
    `[(2 mgh)/(I + mr^2)]^(1/(2))`
    C
    `[(2 mgh)/(I +2 m)]^(1/(2))`
    D
    `sqrt(2 gh)`
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