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WBJEEConsider a disc rotating in the horizontal plane with a constant angular speed...
Consider a disc rotating in the horizontal plane with a constant angular speed ωω about its centre OO. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles PP and QQ are simultaneously projected at an angle towards RR. The velocity of projection is in the yy - zz plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/81/8 rotation, (ii) their range is less than half the disc radius, and (iii) ωω remains constant throughout. Then
Solution:
For 1818 of rotation of disc,
$t=18×2πω=π4ωx-coordinate of P=ωRT=πR4>Rsin45∘t=18×2πω=π4ωx-coordinate of P=ωRT=πR4>Rsin45∘$
To reach the unshaded part, particlePP needs to travel horizontal range >> Rsin45∘≃0.7RRsin45∘≃0.7R
But its range is less than R/2R/2 so it will land on shaded part. QQ is near the origin, its velocity will be nearly along QR so it will land in unshaded part.
$t=18×2πω=π4ωx-coordinate of P=ωRT=πR4>Rsin45∘t=18×2πω=π4ωx-coordinate of P=ωRT=πR4>Rsin45∘$
To reach the unshaded part, particlePP needs to travel horizontal range >> Rsin45∘≃0.7RRsin45∘≃0.7R
But its range is less than R/2R/2 so it will land on shaded part. QQ is near the origin, its velocity will be nearly along QR so it will land in unshaded part.
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