
A meter stick is placed vertically at the origin on a frictionless surface. A gentle push in $ + X$ direction is given to the top most point of the rod when it has fallen completely $X$ coordinate of center of rod is at:
(a) Origin
(b) $ - 0.5m$
(c) $ - 1m$
(d) $ + 0.5m$
Answer
559.2k+ views
Hint: A meter stick is placed in a vertical direction at the origin on a frictionless surface. A gentle push in positive $X$-direction is given to the topmost point of the rod when it has fallen completely, $X$- coordinate of the center of mass of rod
Complete step by step solution:
It looks like there is no external force applied on the rod along the x-axis.
After reading the statement “A gentle push in positive x-direction is given to the topmost point of the rod.”
But we know, if the external force acting on the particle is zero, the position of the center of mass remains the same. It doesn't change.
So, new $X$- coordinate of the center of mass of rod = initial x - coordinate of the center of mass of the rod.
We have the initial center of the mass \[ = {\text{ }}\left( {0,{\text{ }}0.5} \right)\]
That is initial $X$- coordinate of com \[ = {\text{ }}0\]
So, new $X$ - coordinate of com \[ = {\text{ }}0\]
But \[y\]-coordinate has changed because of a gravitational force acting along the \[y\]-axis. And it will be changed from \[0.5\] to zero as shown in the figure.
\[{\text{X }} = {\text{ }}0\] and \[{\text{Y }} = {\text{ }}0\] that is origin
Therefore the new center of mass of the rod is at the origin.
Hence, the correct answer is option (a).
Note:To know the frictionless surface we have to take the example of a skater. If there are no forces acting upon the skater and he is on a frictionless surface, then that means he has no net acceleration. If an object moves with the velocity and if there is no acceleration, then the velocity remains constant.
Complete step by step solution:
It looks like there is no external force applied on the rod along the x-axis.
After reading the statement “A gentle push in positive x-direction is given to the topmost point of the rod.”
But we know, if the external force acting on the particle is zero, the position of the center of mass remains the same. It doesn't change.
So, new $X$- coordinate of the center of mass of rod = initial x - coordinate of the center of mass of the rod.
We have the initial center of the mass \[ = {\text{ }}\left( {0,{\text{ }}0.5} \right)\]
That is initial $X$- coordinate of com \[ = {\text{ }}0\]
So, new $X$ - coordinate of com \[ = {\text{ }}0\]
But \[y\]-coordinate has changed because of a gravitational force acting along the \[y\]-axis. And it will be changed from \[0.5\] to zero as shown in the figure.
\[{\text{X }} = {\text{ }}0\] and \[{\text{Y }} = {\text{ }}0\] that is origin
Therefore the new center of mass of the rod is at the origin.
Hence, the correct answer is option (a).
Note:To know the frictionless surface we have to take the example of a skater. If there are no forces acting upon the skater and he is on a frictionless surface, then that means he has no net acceleration. If an object moves with the velocity and if there is no acceleration, then the velocity remains constant.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

