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JEE Advanced 2022 Paper 1 · Physics · Rotational Motion · easy

A thin uniform spherical shell rolls without slipping on a horizontal floor with centre speed 4 m s^-1 and then, without any loss of speed or…

Numerical answer

A thin uniform spherical shell rolls without slipping on a horizontal floor with centre speed 4 m s−14\ \mathrm{m\,s^{-1}} and then, without any loss of speed or slipping at the junction, rolls up a long incline of angle 30∘30^\circ that is rough enough to prevent slipping. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The maximum distance (in metres) the shell travels along the incline before momentarily stopping is ______ . Give the answer correct to two decimal places.
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Answer: 2.67

For rolling without slipping the total kinetic energy of a thin shell is 12mv2(1+23)=56mv2\tfrac12 mv^2\left(1+\tfrac{2}{3}\right)=\tfrac56 mv^2. Static friction does no net work, so energy conservation gives 56mv2=mg dsin⁡30∘\tfrac56 mv^2=mg\,d\sin30^\circ. Hence d=5v26gsin⁡30∘=5×166×10×0.5=8030=2.67 md=\dfrac{5v^2}{6g\sin30^\circ}=\dfrac{5\times16}{6\times10\times0.5}=\dfrac{80}{30}=2.67\ \mathrm{m}.
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