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प्रश्न
A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Figure). The sphere is struck horizontally at a height h from the floor. Match the following:
| Column I | Column II | |
| (a) h = R/2 | (i) | Sphere rolls without slipping with a constant velocity and no loss of energy. |
| (b) h = R | (ii) | Sphere spins clockwise, loses energy by friction. |
| (c) h = 3R/2 | (iii) | Sphere spins anti-clockwise, loses energy by friction. |
| (d) h = 7R/5 | (iv) | Sphere has only a translational motion, looses energy by friction. |
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उत्तर
| Column I | Column II | |
| (a) h = R/2 | (iii) | Sphere spins anti-clockwise, loses energy by friction. |
| (b) h = R | (iv) | Sphere has only a translational motion, looses energy by friction. |
| (c) h = 3R/2 | (ii) | Sphere spins clockwise, loses energy by friction. |
| (d) h = 7R/5 | (i) | Sphere rolls without slipping with a constant velocity and no loss of energy. |
Explanation:
Mass of the sphere = m
Radius = R
h = height from the floor
The sphere will roll without slipping when ω = V/R
Where v is linear velocity and to is the angular velocity of the sphere.
Now, angular momentum of the sphere is about centre of mass .....[We are applying conservation of angular momentum just before and after struck.]
Then by the law of conservation of angular momentum
`mv(h - R) = I_ω`
`mv(h - R) = 2/5 mR^2 v/R`
`h - R = 2/5 R`
`h = 2/5 R +R = 7/5 R`
Therefore, the sphere rolls without slipping with a constant velocity and no loss of energy. Thus (d) - (i)
Torque due to force `F = τ = (h - R) xx F`
If τ = 0, h – R = 0 and thus h = R
In this case, the sphere will only have a translation motion and slip against the force of friction. Thus (b) - (iv)
For clockwise rotation of the sphere τ > 0
`(h - R) xx F > 0`
Or `h > R`, thus (c) - (ii)
For anti-clockwise rotation `τ < 0`
`(h - R) xx F < 0`
`h < R`,Thus (a) - (iii)
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