JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is
- A \(3.5 \mathrm{~m} / \mathrm{s}^2\)
- B \(0.35 \mathrm{~m} / \mathrm{s}^2\)
- C \(2.5 \mathrm{~m} / \mathrm{s}^2\)
- D \(0.25 \mathrm{~m} / \mathrm{s}^2\)
Answer & Solution
Correct Answer
(A) \(3.5 \mathrm{~m} / \mathrm{s}^2\)
Step-by-step Solution
Detailed explanation
Torque about bottom point \(\begin{aligned} & \mathrm{F} \times 2 \mathrm{r}=\mathrm{I} \alpha \\ & 49 \times 2 \mathrm{r}=\frac{7}{5} \mathrm{mr}^2 \alpha \\ & 14=4 \mathrm{r} \alpha \end{aligned}\) As sphere rolls without slipping…
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