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NEET 2019
Exam
NEET
Year
2019
Subject
Physics
Chapter
Rotational Motion

NEET 2019 · Physics · Rotational Motion

A solid cylinder of mass 2 ~kg and radius 50 ~cm rolls up an...

A solid cylinder of mass 2 kg2 \mathrm{~kg} and radius 50 cm50 \mathrm{~cm} rolls up an inclined plane of angle inclination 30∘30^{\circ}. The centre of mass of cylinder has speed of 4 m/s4 \mathrm{~m} / \mathrm{s}. The distance travelled by the cylinder on the inclined surface will be : (Take g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 )A solid cylinder of mass 2 ~kg and radius 50 ~cm rolls up an inclined plane of angle inclination 30^ . The centre of

Options

  1. A2.2 m2.2 \mathrm{~m}
  2. B1.6 m1.6 \mathrm{~m}
  3. C1.2 m1.2 \mathrm{~m}
  4. D2.4 m2.4 \mathrm{~m}

Correct answer

D. 2.4 m2.4 \mathrm{~m}

Step-by-step solution

When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be KE=12mv2(1+k2R2)\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right) where, m=\mathrm{m}= mass of body, v=\mathrm{v}= velocity and k=\mathrm{k}= radius of gyration A solid cylinder of mass 2 ~kg and radius 50 ~cm rolls up an inclined plane of angle inclination 30^ . The centre of
Given, m=2 kg,θ=30∘,v=4\mathrm{m}=2 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{v}=4 ms−1\mathrm{~ms}^{-1}LetLethbetheheightoftheinclinedplane,thenfromlawofconservationofenergy,be the height of the inclined plane, then from law of conservation of energy,KE=PE12mv2(1+k2R2)=mgh\begin{aligned} \mathrm{KE} & =\mathrm{PE} \\ \frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right) & =\mathrm{mgh} \end{aligned}Substitutingthegivenvaluesintheaboveequation,wegetSubstituting the given values in the above equation, we get12×2×16(1+12)=2×10×h[∵ For cylinder k2R2=12]⇒8×32=10 h⇒h=1.2 m\begin{aligned} & \frac{1}{2} \times 2 \times 16\left(1+\frac{1}{2}\right)=2 \times 10 \times \mathrm{h}\left[\because \text{ For cylinder } \frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{1}{2}\right] \\ & \Rightarrow \quad 8 \times \frac{3}{2}=10 \mathrm{~h} \Rightarrow \mathrm{h}=1.2 \mathrm{~m} \end{aligned}FromtheabovediagramFrom the above diagramsin⁡θ=hx⇒x=hsin⁡θ=1.2sin⁡30∘=1.2×2=2.4 m[∵sin⁡30∘=12]\begin{gathered} \sin \theta=\frac{\mathrm{h}}{\mathrm{x}} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{h}}{\sin \theta}=\frac{1.2}{\sin 30^{\circ}}=1.2 \times 2=2.4 \mathrm{~m}\left[\because \sin 30^{\circ}=\frac{1}{2}\right] \end{gathered}
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