System of Particles & Rotational Motion Test - 4

Body A has mass of 1kg and location (1,2)
Body B has mass of 2kg and location (-1,3)

m_cx_c = m₁x₁ + m₂x₂
(1+2) x_c = (1 * 1) + (2 * -1)
x_c  = -1/3

Similarly,
m_cy_c = m₁y₁ + m₂y₂
(1 + 2) y_c = (1 * 2) + (2 * 3)
y_c= 8/3

Hence, the coordinates of the center of mass are (-1/3, 8/3).

Potter’s wheel is an example of rotary motion. Rotary motion is that kind of motion in which body of the mass moves along a circular path about an axis which remains fixed.

The general motion of a rigid body consists of both, translational and rotational motion. This is much obvious as it is seen all around us.

A rigid body is one in which the distance between all pairs of particles remains fixed.

As a shell explosion is not governed by any external force, we get that no external force acts upon the shell, and hence it remains at rest.

The child is running arbitrarily on a trolley moving with velocity v.
However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley.
This is because the force due to the boy’s motion is purely internal.

Internal forces produce no effect on the motion of the bodies on which they act.

Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.

Given the track is frictionless, therefore as the people inside will move inside the compartment the compartment will itself move opposite to it.
Hence C₁ will move but as their is no external force on the combined system and its initial velocity is also zero. Therefore C₂ won't change.

As the center of mass moves through a distance x, the average force F acting on the man is calculated from :
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F = 

The centre of mass of a body can lie within or outside the body.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.

The internal forces have no effect on the trajectory of the center of mass, and the forces due to explosion are the internal forces. So the center of mass will follow the same parabolic path even after the explosion.

Angular momentum (L) is defined as the distance of the object from a rotation axis multiplied by the linear momentum
L = mv×h

As the particle moves, m; v; and h, all remain unchanged at any point of time
⇒ L = constant

Every point in any rigid rotating object is rotating at the same angular velocity. However, I have seen a few cases where people have used the term "angular velocity" when they really meant tangential velocity, so you do have to be careful.

As the particle is exploded only due to its internal energy.
⇒ net external force during this process is 0 i.e. center mass will not change.

Let the particle while the explosion was above the origin of the coordinate system i.e. just before explosion x_cm =0 and y_cm =0

After the explosion, the Centre of mass will be at x_cm =0 and y_cm =0

Since smaller fragment has fallen on the y-axis.
Let positon of larger fragment be y.

m * y_cm = (m/4 * 15) + (3m/4 * y)
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm

The resultant of all forces, on any system of particles, is zero. Therefore, their centre of mass does not depend upon the forces acting on the particles.

The COM will be at the intersection point of the medians as that is the most symmetric point of this symmetric system.

When force acting upon the body results zero, the resulting acceleration due to net force applied is also zero, and hence by the law of inertia the motion of the body either at rest or constant velocity wont change.

Velocity of heavier block (v₂) = 14m/s
Velocity of lighter block (v₁) = 0m/s
Velocity of centre of mass,

Using, 

Let origin be the centre of mass of the body which is represented by C i.e. r = 0
⟹ m₁r₁ + m₂​r₂ + m₃r₃ = 0

Now net moment about C,
τC = (m₁g) * r₁ + (m₂g) * r₂ + (m₃g) * r₃
τC = g [m₁r₁ + m₂​r₂ + m₃r₃] = 0

Hence, the sum of moment of masses of all the particles in the system about COM is always Zero. 

The distance of centre of mass from carbon atom, 

• For an object to have the x-coordinate of its centre of mass, there should be particles on both the negative and positive coordinates of the plane for the sum to be zero.
• However, for 1-dimensional particles that all lie on the y-axis symmetric about the x-axis, the x-coordinate is zero – which is neither positive nor negative, i.e., non-negative.

Sum of masses = 1 + 1.5 + 2 + M = 4.5 + M
x-coordinate;
(1*2 + 1.5*1 + 2*2 – M)/(4.5 + M) = 1
4.5 + M = 7.5 – M
2M = 3
M = 1.5 kg.

• Distance between first and last sphere = R + 2R + R = 4R
• Since the spheres are identical and lie in a straight line, the centre of mass will lie exactly in the middle.
• Hence the centre of mass lies at a distance of 2R from the centre of the first sphere.

• The centre of mass of a bangle lies at its geometric centre which does not lie on or in the object.
• Hence the centre of mass of a bangle lies outside the body.

Sum of masses = 2 + 1 = 3
x-coordinate;
(2*1 + 1*2)/3 = 4/3
y-coordinate;
(2*0 + 1*2)/3 = 2/3
z-coordinate;
(2*1 + 1*0)/3 = 2/3.