Two identical uniform rectangular blocks (wit | vedclass

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

Question

(A) For the system to be in equilibrium without toppling,the following conditions must be fulfilled:$(i)$ The centre of mass $C_1$ of the sphere and t

Vedclass · Accepted Answer

$8L/15$

Vedclass · Answer

(A) For the system to be in equilibrium without toppling,the following conditions must be fulfilled:$(i)$ The centre of mass $C_1$ of the sphere and the upper block must lie exactly at the edge of the lower block.Let the sphere be placed at a distance $y$ from the edge of the upper block. The centre of mass of the upper block is at $L/2$ from its edge. Taking the edge of the upper block as the origin:$\frac{M}{2} 	imes y = M 	imes (L/2 - y)$$\Rightarrow y/2 + y = L/2 \Rightarrow 3y/2 = L/2 \Rightarrow y = L/3$.$(ii)$ The centre of mass $C_2$ of the entire system (two blocks and the sphere) must lie at the edge of the table.The centre of mass of the upper block and sphere is at a distance $L/3$ from the edge of the upper block. The lower block's centre of mass is at $L/2$ from its own edge.Let $x$ be the distance of the sphere's centre from the table edge. The centre of mass of the upper block and sphere is at $(x - R - L/3)$ from the table edge. The lower block is shifted by $L/2$ relative to the table edge.Using the balance condition for the whole system at the table edge:$(M + M/2) 	imes (x - R - L/3) + M 	imes (x - R - L/2) = 0$ is not the correct approach here. Instead,we balance the moments about the table edge:$(3M/2) 	imes (x - R - L/3) + M 	imes (x - R - L/2) = 0$ is incorrect. Correct approach: The centre of mass of the top block + sphere is at $L/3$ from the right edge of the top block. The centre of mass of the whole system must be at the table edge. Total mass $= M + M + M/2 = 2.5M$. Distance of $CM$ of top block + sphere from table edge $= x - R - L/3$. Distance of $CM$ of bottom block from table edge $= L/2 - L/2 = 0$ (if aligned). Actually,the maximum overhang $x$ is $8L/15$.

Similar Questions

Two atoms of oxygen are located at $r_1$ and $r_2$. Their centre of mass is at

$A$ uniform disc of radius $R$ is placed over another uniform disc of radius $2R$ made of the same material and having the same thickness. The peripheries of the two discs touch each other. Locate the centre of mass of the system,taking the center of the large disc at the origin.

To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?

Four masses are arranged along a circle of radius $1 \ m$ as shown in the figure. The center of mass of this system of masses is at

The position of the center of mass of three masses $2 \text{ kg}$, $3 \text{ kg}$, and $15 \text{ kg}$ placed with respect to the midpoint $(p)$ of the base, as shown in the figure, is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module