A rod has non-uniformly distributed mass M over its length L . Its linear mass density variation with distance x from the left end is shown in the figure. From this information, we can conclude that the centre of mass of the rod,

Question

A rod has non-uniformly distributed mass M over its length L . Its linear mass density variation with distance x from the left end is shown in the figure. From this information, we can conclude that the centre of mass of the rod, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-3vcsvhfmjxpviums.jpg" /> (A) is on the left of its mid-point. (B) is on the right of its mid-point. (C) coincides with its mid-point. (D) information is insufficient.

Tardigrade Admin · Accepted Answer

The area under λ versus x curve gives mass. The area of I> area of II . It means more mass is on the left side. So, the centre of mass would be in the left half. <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-3vcsvhfmjxpvy0zc.jpg" />

Q. A rod has non-uniformly distributed mass $M$ over its length $L$ . Its linear mass density variation with distance $x$ from the left end is shown in the figure. From this information, we can conclude that the centre of mass of the rod,

is on the left of its mid-point.

is on the right of its mid-point.

coincides with its mid-point.

information is insufficient.

The area under $λ$ versus $x$ curve gives mass.
The area of $I >$ area of $II$ . It means more mass is on the left side. So, the centre of mass would be in the left half.

Q. A rod has non-uniformly distributed mass M over its length L . Its linear mass density variation with distance x from the left end is shown in the figure. From this information, we can conclude that the centre of mass of the rod,