A block of 200 g mass moves with a uniform s | vedclass

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(D) The normal force $N$ provides the necessary centripetal force for circular motion: $N = m \omega^2 R$. Given $m = 200 \ g = 0.2 \ kg$,$R = 20 \ cm

Vedclass · Accepted Answer

$9.859 	imes 10^{-4} \ N$

Vedclass · Answer

(D) The normal force $N$ provides the necessary centripetal force for circular motion: $N = m \omega^2 R$. Given $m = 200 \ g = 0.2 \ kg$,$R = 20 \ cm = 0.2 \ m$,and $T = 40 \ s$. The angular velocity is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{40} = \frac{\pi}{20} \ rad/s$. Substituting the values: $N = m \left( \frac{2 \pi}{T} ight)^2 R = 0.2 	imes \left( \frac{2 \pi}{40} ight)^2 	imes 0.2$. $N = 0.2 	imes \frac{4 \pi^2}{1600} 	imes 0.2 = 0.04 	imes \frac{\pi^2}{400} = 10^{-4} 	imes \pi^2$. Using $\pi^2 \approx 9.859$,we get $N = 9.859 	imes 10^{-4} \ N$.

$A$ block of $200 \ g$ mass moves with a uniform speed in a horizontal circular groove,with vertical side walls of radius $20 \ cm$. 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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