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Rotational Motion question

2019 · 9 Apr · Shift 1 · Q54
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Rotational Motion question

2019 · 9 Apr · Shift 1 · Q54

JEE MainPhysicsRotational MotionMCQ+4 / −1
The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane. : (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radius R/4 . If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is :
  1. A
    20 : 15 : 14
  2. B
    4 : 3 : 2
  3. C
    2 : 3 : 4
  4. D
    10 : 15 : 7
View written solutionFree

Correct answer: A

  1. Use conservation of mechanical energy

When a rigid body rolls without slipping up the incline, its initial total kinetic energy converts into gravitational potential energy at the highest point.

So,

Ktrans+Krot=mghmax⁡K_{\text{trans}} + K_{\text{rot}} = mgh_{\max}Ktrans​+Krot​=mghmax​

Given that the speed of the centre of mass at the bottom is same in all three cases, let it be vvv.

Also, for rolling without slipping,

ω=vr\omega = \frac{v}{r}ω=rv​

where rrr is the radius of the body.

Thus,

mgh=12mv2+12Iω2mgh = \frac12 mv^2 + \frac12 I\omega^2mgh=21​mv2+21​Iω2 =12mv2+12I(vr)2= \frac12 mv^2 + \frac12 I\left(\frac{v}{r}\right)^2=21​mv2+21​I(rv​)2

Hence,

h∝12mv2(1+Imr2)h \propto \frac12 mv^2\left(1 + \frac{I}{mr^2}\right)h∝21​mv2(1+mr2I​)

Since mmm and vvv are same-factor common for comparison,

h∝1+Imr2h \propto 1 + \frac{I}{mr^2}h∝1+mr2I​

So we only need the factor 1+Imr21 + \frac{I}{mr^2}1+mr2I​ for each body.

  1. For the ring

For a ring,

I=mr2I = mr^2I=mr2

Therefore,

1+Imr2=1+1=21 + \frac{I}{mr^2} = 1 + 1 = 21+mr2I​=1+1=2

So,

hring∝2h_{\text{ring}} \propto 2hring​∝2
  1. For the solid cylinder

For a solid cylinder,

I=12mr2I = \frac12 mr^2I=21​mr2

Therefore,

1+Imr2=1+12=321 + \frac{I}{mr^2} = 1 + \frac12 = \frac321+mr2I​=1+21​=23​

So,

hcyl∝32h_{\text{cyl}} \propto \frac32hcyl​∝23​
  1. For the solid sphere

For a solid sphere,

I=25mr2I = \frac{2}{5}mr^2I=52​mr2

Therefore,

1+Imr2=1+25=751 + \frac{I}{mr^2} = 1 + \frac{2}{5} = \frac751+mr2I​=1+52​=57​

So,

hsphere∝75h_{\text{sphere}} \propto \frac75hsphere​∝57​
  1. Form the ratio

Therefore,

hring:hcyl:hsphere=2:32:75h_{\text{ring}} : h_{\text{cyl}} : h_{\text{sphere}} = 2 : \frac32 : \frac75hring​:hcyl​:hsphere​=2:23​:57​

Multiply by 101010 to clear fractions:

20:15:1420 : 15 : 1420:15:14
  1. Check role of given radii

The radii are RRR, R/2R/2R/2, and R/4R/4R/4, but they do not affect the result because for each standard body,

Imr2\frac{I}{mr^2}mr2I​

is independent of radius. Hence the given radii are irrelevant to the final ratio.

  1. Evaluate options
  • A: 20:15:1420:15:1420:15:14 ✅
  • B: 4:3:24:3:24:3:2 ❌
  • C: 2:3:42:3:42:3:4 ❌
  • D: 10:15:710:15:710:15:7 ❌

So the correct option is A.

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