Question:
The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point ($p$) of normal bisector, as shown in the figure is _________
The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point ($p$) of normal bisector, as shown in the figure is _________
Show Hint
Define a coordinate system with point $p$ as origin. Calculate the altitude of the triangle to find the y-coordinates of the masses, and use the side lengths to find the x-coordinates.
Updated On: Apr 9, 2026
- $(\frac{\sqrt{3}}{4}, 1.25)$
- $(\frac{\sqrt{3}}{4}, 1.0)$
- (0,0)
- (1.25, 0)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
To find the center of mass of the given system, we first need to establish a coordinate system centered at point $p$.
Step 1: Understand the geometry.
The masses are placed at the vertices of an isosceles triangle with side lengths of 10 m and a vertex angle of $120^\circ$ where the 15 kg mass is located. The "normal bisector" of the base is the altitude of this triangle. Point $p$ is the midpoint of this altitude.
Step 2: Calculate the dimensions of the triangle.
Let the altitude (from the $120^\circ$ vertex to the base) be $H$. In the right triangle formed by the altitude and a side:
The angle at the top is bisected, so it is $60^\circ$.
Altitude, $H = 10 \cos(60^\circ) = 10 \times 0.5 = 5$ m.
Half-base width, $W = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m.
Step 3: Define coordinates relative to point $p(0,0)$.
Since $p$ is the midpoint of the altitude $H=5$ m, the top vertex is at $+2.5$ m on the y-axis and the base is at $-2.5$ m on the y-axis.
- Mass $m_1 = 2$ kg (left vertex): $x_1 = -5\sqrt{3}, y_1 = -2.5$
- Mass $m_2 = 3$ kg (right vertex): $x_2 = 5\sqrt{3}, y_2 = -2.5$
- Mass $m_3 = 15$ kg (top vertex): $x_3 = 0, y_3 = 2.5$
Step 4: Use the Center of Mass formula.
Total mass $M_{total} = 2 + 3 + 15 = 20$ kg.
$$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M_{total}} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20}$$
$$X_{cm} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$$
$$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M_{total}} = \frac{2(-2.5) + 3(-2.5) + 15(2.5)}{20}$$
$$Y_{cm} = \frac{-5 - 7.5 + 37.5}{20} = \frac{25}{20} = 1.25$$
The center of mass coordinates relative to point $p$ are $(\frac{\sqrt{3}}{4}, 1.25)$. This corresponds to option 1.
Step 1: Understand the geometry.
The masses are placed at the vertices of an isosceles triangle with side lengths of 10 m and a vertex angle of $120^\circ$ where the 15 kg mass is located. The "normal bisector" of the base is the altitude of this triangle. Point $p$ is the midpoint of this altitude.
Step 2: Calculate the dimensions of the triangle.
Let the altitude (from the $120^\circ$ vertex to the base) be $H$. In the right triangle formed by the altitude and a side:
The angle at the top is bisected, so it is $60^\circ$.
Altitude, $H = 10 \cos(60^\circ) = 10 \times 0.5 = 5$ m.
Half-base width, $W = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m.
Step 3: Define coordinates relative to point $p(0,0)$.
Since $p$ is the midpoint of the altitude $H=5$ m, the top vertex is at $+2.5$ m on the y-axis and the base is at $-2.5$ m on the y-axis.
- Mass $m_1 = 2$ kg (left vertex): $x_1 = -5\sqrt{3}, y_1 = -2.5$
- Mass $m_2 = 3$ kg (right vertex): $x_2 = 5\sqrt{3}, y_2 = -2.5$
- Mass $m_3 = 15$ kg (top vertex): $x_3 = 0, y_3 = 2.5$
Step 4: Use the Center of Mass formula.
Total mass $M_{total} = 2 + 3 + 15 = 20$ kg.
$$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M_{total}} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20}$$
$$X_{cm} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$$
$$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M_{total}} = \frac{2(-2.5) + 3(-2.5) + 15(2.5)}{20}$$
$$Y_{cm} = \frac{-5 - 7.5 + 37.5}{20} = \frac{25}{20} = 1.25$$
The center of mass coordinates relative to point $p$ are $(\frac{\sqrt{3}}{4}, 1.25)$. This corresponds to option 1.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main Centre of mass Questions
- Two solid spheres each of mass 2 kg and radius 75 cm are arranged as shown. Find MOI of the system about the given axis.
- Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure
The tensions T1 and T2 in the string are respectively: Find the value of m if \(M = 10\) \(kg\). All the surfaces are rough.
As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball $Q$ after collision will be :$\left( g =10 \,m / s ^2\right)$
- Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3 m$ each Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{ x } m$ The value of $x$ is
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions