A uniform rod of mass M and length L is place | vedclass

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

Question

(A) The motion of the centre of mass of a system of particles is governed by the net external force acting on the system.According to Newton's second

Vedclass · Accepted Answer

$F/M$

Vedclass · Answer

(A) The motion of the centre of mass of a system of particles is governed by the net external force acting on the system.According to Newton's second law for the centre of mass,$F_{	ext{ext}} = M_{	ext{total}} \cdot a_{	ext{cm}}$.Here,the total mass of the rod is $M$ and the net external force applied is $F$.Therefore,$F = M \cdot a_{	ext{cm}}$.Solving for the acceleration of the centre of mass,we get $a_{	ext{cm}} = F/M$.Thus,the correct option is $A$.

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

Similar Questions

The maximum number of sub-levels,orbitals and electrons in $N$ shell of an atom are respectively

If two sides of a triangle are represented by $3x^2-5xy+2y^2=0$ and its orthocentre is $(2,1)$,then the equation of the third side is

The cell wall of which of the following is made of peptidoglycan?

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, Wb/m^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is:

The shadow of a tower standing on a level ground is found to be $60 \ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is

Vedclass Test Series

Exam Paper Generator

Online Exam Module