A 4 kg mass moves under the influence of a forcevecmathrmF = (4mathrmt^3hatmathrmi - 3mathrmthatmathrmj)text N$\vec{\mathrm{F}} = (4\mathrm{t}^3\hat{\mathrm{i}} - 3\mathrm{t}\hat{\mathrm{j}})\text{ N}$ where t is the time in second. If mass starts from origin at t = 0$t = 0$, the velocity and position after t = 2texts$t = 2\text{s}$ will be:
A coin is placed on a disc. The coefficient of friction between the coin and the disc is mu$\mu$. If the distance of the coin from the center of the disc is r$r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
A.fracmu gr$\frac{\mu g}{r}$
B.sqrtfracrmu g$\sqrt{\frac{r}{\mu g}}$
C.sqrtfracmu gr$\sqrt{\frac{\mu g}{r}}$
D.fracmusqrtrg$\frac{\mu}{\sqrt{rg}}$
Solution
### Related Formula
f_s leq mu_s N$$f_s \leq \mu_s N$$F_c = mromega^2$$F_c = mr\omega^2$$
### Core Logic
Circular Motion Friction diagram for Q47 - JEE Main 2024 MorningCircular Motion Friction diagram for Q47 - JEE Main 2024 Morning
To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion.
f = momega^2 r$f = m\omega^2 r$
The normal force on the flat disc is N = mg$N = mg$.
The maximum static friction is f_textmax = mu N = mu mg$f_{\text{max}} = \mu N = \mu mg$.
For no slipping:
m r omega^2 leq mu mg$$m r \omega^2 \leq \mu mg$$omega^2 leq fracmu gr$$\omega^2 \leq \frac{\mu g}{r}$$omega_textmax = sqrtfracmu gr$$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$$
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Laws Of Motion
More Laws of Motion Questions — jee_main_2026_21_jan_morning
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