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Rotational Motion question

2019 · 8 Apr · Shift 2 · Q68
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Rotational Motion question

2019 · 8 Apr · Shift 2 · Q68

JEE MainPhysicsRotational MotionMCQ+4 / −1
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio hsph/hcyl is given by :- JEE Main 2019 (Online) 8th April Evening Slot Physics - Rotational Motion Question 173 English
  1. A
    1
  2. B
    14/15
  3. C
    4/5
  4. D
    2/5\sqrt55​
View written solutionFree

Correct answer: B

  1. Use conservation of mechanical energy

Since both bodies roll without slipping and there is no energy loss, their initial total kinetic energy converts into gravitational potential energy at the highest point.

Thus, mgh=Ktrans+Krotmgh = K_{\text{trans}} + K_{\text{rot}}mgh=Ktrans​+Krot​

where Ktrans=12mv2,Krot=12Iω2K_{\text{trans}} = \frac12 mv^2, \qquad K_{\text{rot}} = \frac12 I\omega^2Ktrans​=21​mv2,Krot​=21​Iω2

Also, for rolling without slipping, v=ωR⇒ω=vRv = \omega R \Rightarrow \omega = \frac{v}{R}v=ωR⇒ω=Rv​

So, mgh=12mv2+12Iv2R2mgh = \frac12 mv^2 + \frac12 I\frac{v^2}{R^2}mgh=21​mv2+21​IR2v2​

Hence, h=v22g(1+ImR2)h = \frac{v^2}{2g}\left(1 + \frac{I}{mR^2}\right)h=2gv2​(1+mR2I​)

  1. For the solid sphere

Moment of inertia of a solid sphere about its center: Isph=25mR2I_{\text{sph}} = \frac{2}{5}mR^2Isph​=52​mR2

Therefore,

= \frac{v^2}{2g}\cdot \frac{7}{5}$$ --- 3. **For the solid cylinder** Moment of inertia of a solid cylinder about its axis: $$I_{\text{cyl}} = \frac{1}{2}mR^2$$ Therefore, $$h_{\text{cyl}} = \frac{v^2}{2g}\left(1 + \frac{1}{2}\right) = \frac{v^2}{2g}\cdot \frac{3}{2}$$ --- 4. **Find the ratio** $$\frac{h_{\text{sph}}}{h_{\text{cyl}}} = \frac{\frac{7}{5}}{\frac{3}{2}} = \frac{7}{5}\cdot \frac{2}{3} = \frac{14}{15}$$ --- 5. **Check options** - A: $1$ ❌ - B: $\frac{14}{15}$ ✅ - C: $\frac{4}{5}$ ❌ - D: $\frac{2}{\sqrt{5}}$ ❌ So the correct option is **B**.
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