A particle of mass m is released from a heigh | vedclass

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

Question

(D) By conservation of energy,the velocity $v$ of the particle at an angle $	heta$ is given by: $mgH = mgR(1 + \cos 	heta) + \frac{1}{2}mv^2$,which

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) By conservation of energy,the velocity $v$ of the particle at an angle $	heta$ is given by: $mgH = mgR(1 + \cos 	heta) + \frac{1}{2}mv^2$,which simplifies to $v^2 = 2g(H - R - R \cos 	heta)$.Applying Newton's second law in the radial direction: $N - mg \cos 	heta = \frac{mv^2}{R}$.Substituting $v^2$: $N = mg \cos 	heta + \frac{m}{R}[2g(H - R - R \cos 	heta)] = mg \cos 	heta + \frac{2mgH}{R} - 2mg - 2mg \cos 	heta = \frac{2mgH}{R} - 2mg - mg \cos 	heta$.At $	heta = 0$,$N = \frac{2mgH}{R} - 3mg$,which is the maximum value.At $	heta = \pi$,$N = \frac{2mgH}{R} - mg$,which is the minimum value (if $H > 5R/2$,$N > 0$ at the top).Thus,all statements are correct.

Similar Questions

$A$ bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies:

$ABCDE$ is a channel in the vertical plane,part $BCDE$ being circular with radius $r$. $A$ block is released from $A$ and slides without friction and without rolling. The block will complete the loop if $h$ is

$A$ particle is whirled in a vertical circle of radius $1.0 \ m$ using a string with one end fixed. If the ratio of maximum and minimum tension in the string is $\frac{5}{3}$,the minimum velocity of the particle during circular motion is ......... $ms^{-1}$.

$A$ $1 \,kg$ stone at the end of a $1 \,m$ long string is whirled in a vertical circle at a constant speed of $4 \,m/s$. The tension in the string is $6 \,N$ when the stone is at:

$A$ particle starts sliding from the top of a sphere of diameter $42 \, m$. At what height from the bottom of the sphere will the particle lose contact with the sphere?

Vedclass Test Series

Exam Paper Generator

Online Exam Module