Question:
A block of $200 \mathrm{~g}$ mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius $20 \mathrm{~cm}$. If the block takes $40 \mathrm{~s}$ to complete one round, the normal force by the side walls of the groove is:
Correct Option: , 4
Solution:
(4)
$\mathrm{N}=\mathrm{m} \omega^{2} \mathrm{R}$
$\mathrm{N}=\mathrm{m}\left[\frac{4 \pi^{2}}{\mathrm{~T}^{2}}\right] \mathrm{R}$
Given $\mathrm{m}=0.2 \mathrm{~kg}, \mathrm{~T}=40 \mathrm{~S}, \mathrm{R}=0.2 \mathrm{~m}$
Put values in equation (1)
$\mathrm{N}=9.859 \times 10^{-4} \mathrm{~N}$