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MHT CET 2024 (10 May, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2024 (10 May, Shift 1)

Q28MHT CET 2024 · 10 MAY · SHIFT 1Medium

Rolling Motion

A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is

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Answer: (d)

  1. Rolling KE =12mv2(1+k2/R2)= \frac{1}{2}mv^2(1 + k^2/R^2).
  2. Ring: k2/R2=1⇒KE=mv2=6k^2/R^2 = 1 \Rightarrow KE = mv^2 = 6 J.
  3. Disc: k2/R2=12⇒KE=34mv2=34×6=92k^2/R^2 = \frac{1}{2} \Rightarrow KE = \frac{3}{4}mv^2 = \frac{3}{4}\times 6 = \frac{9}{2} J.
  4. So the answer is (d).
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