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

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$6.28 \times 10^{- 3} N$

$9.869 \times 10^{- 2} N$

$9.869 \times 10^{- 4} N$

$0.0314 N$

answer is A.

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Detailed Solution

$A n g u l a r v e l o c i t y = ω = \frac{2 π}{T} = \frac{2 π}{40} N o r m a l r e a c t i o n p r o v i d e s c e n t r i p e t a l f o r c e . N = m r ω^{2} = \frac{200}{1000} \times \frac{20}{100} \times {(\frac{2 π}{40})}^{2} = 9.869 \times 10^{- 4} N$

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