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JEE Advanced 2017Paper 2
Exam
JEE Advanced
Year
2017
Paper
Paper 2
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2017 · Paper 2 · Physics · Rotational Motion

Consider regular polygons with number of sides n = 3,4, 5 as shown...

Consider regular polygons with number of sides n=3,4,5…n = 3,4, 5 \ldots as shown in the figure. The center of mass of all the polygons is at height hh from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ\Delta . Then, Δ\Delta depends on nn and hh as
Consider regular polygons with number of sides n = 3,4, 5 as shown in the figure. The center of mass of all the

Options

  1. AΔ=hsin⁡2(πn)\Delta = h\sin^{2} \left(\dfrac{\pi}{n}\right)
  2. BΔ=hsin(2πn)\Delta = hsin \left(\dfrac{2\pi}{n}\right)
  3. CΔ = h\left(\dfrac{1}{\cos \left(\dfrac{π}{n}\right)} - 1\right)
  4. DΔ=htan⁡2(π2n)\Delta = h\tan^{2} \left(\dfrac{\pi}{2n}\right)

Correct answer

C. Δ = h\left(\dfrac{1}{\cos \left(\dfrac{π}{n}\right)} - 1\right)

Step-by-step solution

Consider regular polygons with number of sides n = 3,4, 5 as shown in the figure. The center of mass of all the

OA=hOA = h

OB=hcosπnOB = \dfrac{h}{cos \dfrac{\pi}{n}}

Initial height of COM=hCOM = h

Final height of COM=hcos⁡(πn)COM = \dfrac{h}{\cos \left(\dfrac{\pi}{n}\right)}

∴Δ=hcosπn−h∴ \Delta = \dfrac{h}{cos \dfrac{\pi}{n}} - h

Δ=h[1cosπn−1]\Delta = h\left[\dfrac{1}{cos \dfrac{\pi}{n}} - 1\right]

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