Question

A block of $200\text{g}$ mass moves with a uniform speed in a horizontal circular groove of radius $20\text{cm}$, with vertical side walls. If the block takes $40\text{s}$ to complete one round, the normal force by the side walls of the groove is:

• A. $6.28\times {10}^{-3}\text{N}$
• B. $0.0314\text{N}$
• C. $9.859\times {10}^{-2}\text{N}$
• D. $9.859\times {10}^{-4}\text{N}$

Answer 1

The correct option is D $9.859\times {10}^{-4}\text{N}$

Normal force will provide the necessary centripetal force.
$\Rightarrow N=m{\omega }^{2}R$
Also, $\omega =\frac{2\pi }{T}$
$\Rightarrow N=\left(0.2\right)\left(\frac{4{\pi }^2}{T^2}\right)\left(0.2\right)$
$\Rightarrow N=0.2\times \frac{4\times {\left(3.14\right)}^2}{{\left(40\right)}^2}\times 0.2$
$\therefore N=9.859\times {10}^{-4}\text{N}$

Hence, option (d) is correct.