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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The normal force $N$ provides the necessary centripetal force for circular motion.$N = F_c = m \omega^2 R$Since $\omega = \frac{2\pi}{T}$,we have

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The normal force $N$ provides the necessary centripetal force for circular motion.$N = F_c = m \omega^2 R$Since $\omega = \frac{2\pi}{T}$,we have $N = m \left( \frac{2\pi}{T} ight)^2 R = m \frac{4\pi^2}{T^2} R$.Given values: $m = 200\, g = 0.2\, kg$,$R = 20\, cm = 0.2\, m$,and $T = 40\, s$.Substituting these values into the formula:$N = 0.2 	imes \frac{4 	imes (3.14159)^2}{(40)^2} 	imes 0.2$$N = 0.2 	imes \frac{4 	imes 9.8696}{1600} 	imes 0.2$$N = 0.04 	imes \frac{39.4784}{1600}$$N = 0.04 	imes 0.024674 = 9.8696 	imes 10^{-4}\, N$.Rounding to the nearest option,we get $N \approx 9.859 	imes 10^{-4}\, N$.

$A$ 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

Similar Questions

$A$ particle of mass $m$ is rotating in a plane in a circular path of radius $r$. Its angular momentum is $L$. The centripetal force acting on the particle is

$A$ train is moving towards north. At one place it turns towards north-east. Here,we observe that:

$A$ $500 \,kg$ car takes a round turn of radius $50 \,m$ with a velocity of $36 \,km/h$. The centripetal force is (in $\,N$)

$A$ particle is moving with constant speed along a uniform helical path $(a)$ and a spiral path $(b)$ separately,as shown in the figure.

$A$ motorcycle is moving on an overbridge of radius $R$. The driver maintains a constant speed $v$. As the motorcycle is ascending on the overbridge,the normal force on it

Vedclass Test Series

Exam Paper Generator

Online Exam Module