Question

A small block of mass m slides along a smooth frictionless track connected with a circular track of radius r if it starts from rest at p

Answer 1

A small block of mass m slides along a smooth circular track of radius R as shown in the figure. If it starts from rest at P, what is the resultant force acting on it at Q?

A. $\mathrm{F}=\sqrt{35 \mathrm{mg}}$

B. $\mathrm{F}=\sqrt{55} \mathrm{mg}$

C. $\mathrm{F}=\sqrt{65 \mathrm{mg}}$

D. $\mathrm{F}=\sqrt{84} \mathrm{mg}$

Answer:

Total force on the block at Q is $\mathrm{F}=\sqrt{\mathrm{N}^{2}+\mathrm{mg}}=\sqrt{(8 \mathrm{mg})^{2}+(\mathrm{mg})^{2}}=\sqrt{65 \mathrm{mg}}$

Explanation:

Let the velocity of the block at point Q be v.

By Work-energy theorem : $\mathrm{W}=\Delta \mathrm{K} . \mathrm{E}$

$\therefore \mathrm{mg}(5 \mathrm{R}-\mathrm{R})=\frac{1}{2} \mathrm{mv}^{2}-0$

$\Longrightarrow \mathrm{mv}^{2}=8 \mathrm{mg}$

Using circular motion equation: $\mathrm{N}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}$

$\therefore \mathrm{N}=\frac{8 \mathrm{mgR}}{\mathrm{R}}=8 \mathrm{mg}$

Total force on the block at Q is

$\mathrm{F}=\sqrt{\mathrm{N}^{2}+\mathrm{mg}}=\sqrt{(8 \mathrm{mg})^{2}+(\mathrm{mg})^{2}}=\sqrt{65 \mathrm{mg}}$

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