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  • A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the centre. If the angular the velocity of the table is halved, it will just slip when placed a

A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the centre. If the angular
the velocity of the table is halved, it will just slip when placed at a distance of ______ from the centre :

Option: 1

\mathrm{1\: cm}


Option: 2

4\mathrm{cm}


Option: 3

2\mathrm{cm}


Option: 4

1\mathrm{cm}


Answers (1)

best_answer

When the coin is just on the verge of slipping, it just overcomes the force of static friction (Fr). which acts as a centripetal force for being that coin in the circular motion.

If we halved the \omega, then 'r' will also self-adjust to a new  r_1, since the static friction will remain same. 

\begin{aligned} & \mathrm{fr}=\mathrm{m} \omega^2 \mathrm{r} \\ & \mu \mathrm{mg}=\mathrm{m} \omega^2 \mathrm{r}=\mathrm{const} . \\ & \omega^2 \mathrm{r}=\mathrm{const} \\ & \omega_1^2 \mathrm{r}_1=\omega_2^2 \mathrm{r}_2 \\ & \omega^2(1)=\left(\frac{\omega}{2}\right)^2 \mathrm{r}_2 \\ & \mathrm{r}_2=4 \mathrm{~cm} \end{aligned}

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HARSH KANKARIA

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