A uniform wire is bent into a U shape with si | vedclass

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

Question

(D) Let the mass of a wire of length $l$ be $m$.For segment $AB$ (length $l$): Mass $m_1 = m$,Center of mass $\vec{r}_1 = \left( \frac{l}{2}, 0 \right

Vedclass · Accepted Answer

$\left( \frac{l}{4}, l \right)$

Vedclass · Answer

(D) Let the mass of a wire of length $l$ be $m$.For segment $AB$ (length $l$): Mass $m_1 = m$,Center of mass $\vec{r}_1 = \left( \frac{l}{2}, 0 \right)$.For segment $BC$ (length $2l$): Mass $m_2 = 2m$,Center of mass $\vec{r}_2 = (0, l)$.For segment $CD$ (length $l$): Mass $m_3 = m$,Center of mass $\vec{r}_3 = \left( \frac{l}{2}, 2l \right)$.The coordinates of the center of mass $(X_{cm}, Y_{cm})$ are given by:$X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{m(\frac{l}{2}) + 2m(0) + m(\frac{l}{2})}{m + 2m + m} = \frac{m(l)}{4m} = \frac{l}{4}$.$Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{m(0) + 2m(l) + m(2l)}{m + 2m + m} = \frac{4ml}{4m} = l$.Thus,the center of mass is $\left( \frac{l}{4}, l \right)$.

$A$ uniform wire is bent into a $U$ shape with side lengths $l, 2l,$ and $l$. The coordinates of the center of mass of each segment are shown in the figure. The coordinates of the center of mass of the system are:

Similar Questions

The distance of the centre of mass of a solid uniform cone from its vertex is $z_0$. If the radius of its base is $R$ and its height is $h$,then $z_0$ is equal to:

If the linear density of a rod of length $3 \ m$ varies as $\lambda = 2 + x$,then the position of the centre of mass of the rod is

$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

How can a general body be treated in mechanics?

Vedclass Test Series

Exam Paper Generator

Online Exam Module