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Rotational Motion question

2024 · 29 Jan · Shift 1 · Q82
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  5. /2024 · 29 Jan · Shift 1 · Q82

Rotational Motion question

2024 · 29 Jan · Shift 1 · Q82

JEE MainPhysicsRotational MotionNumerical+4 / −1
A cylinder is rolling down on an inclined plane of inclination 60∘60^{\circ}60∘. It's acceleration during rolling down will be x3m/s2\frac{x}{\sqrt{3}} m / s^23​x​m/s2, where x=x=x=‾\underline{\hspace{2cm}}​ (use g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2g=10 m/s2).
Numerical answer
View written solutionFree

Correct answer: 10

  1. Acceleration of a body rolling without slipping down an incline

For a rigid body rolling down an incline,

a=gsin⁡θ1+ImR2a = \frac{g\sin\theta}{1+\dfrac{I}{mR^2}}a=1+mR2I​gsinθ​

where:

  • III = moment of inertia about center,
  • mmm = mass,
  • RRR = radius.
  1. For a solid cylinder

The moment of inertia is

I=12mR2I = \frac{1}{2}mR^2I=21​mR2

So,

ImR2=12\frac{I}{mR^2} = \frac{1}{2}mR2I​=21​
  1. Substitute into the formula
a=gsin⁡θ1+12=gsin⁡θ32=23gsin⁡θa = \frac{g\sin\theta}{1+\frac{1}{2}} = \frac{g\sin\theta}{\frac{3}{2}} = \frac{2}{3}g\sin\thetaa=1+21​gsinθ​=23​gsinθ​=32​gsinθ
  1. Use θ=60∘\theta=60^\circθ=60∘ and g=10 m/s2g=10\,\text{m/s}^2g=10m/s2
a=23⋅10⋅sin⁡60∘a = \frac{2}{3}\cdot 10 \cdot \sin 60^\circa=32​⋅10⋅sin60∘

Since

sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}sin60∘=23​​

therefore,

a=203⋅32=1033a = \frac{20}{3}\cdot \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3}a=320​⋅23​​=3103​​

Now,

1033=103\frac{10\sqrt{3}}{3} = \frac{10}{\sqrt{3}}3103​​=3​10​

So the acceleration is

a=103 m/s2a = \frac{10}{\sqrt{3}}\,\text{m/s}^2a=3​10​m/s2
  1. Compare with given form

Given,

a=x3 m/s2a = \frac{x}{\sqrt{3}}\,\text{m/s}^2a=3​x​m/s2

Thus,

x=10x = 10x=10

Final Answer:

10\boxed{10}10​
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