Telangana TSBIE TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.

## TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion

Very Short Answer Type Questions

Question 1.

Is it necessary that a mass should be present at the centre of mass of any system? [AP May. ’16; May ’14]

Answer:

No. It is not necessary to present some mass at centre of mass of the system.

Ex: At the centre of ring (or) bangle, there is no mass present at centre of mass.

Question 2.

What is the difference in the positions of a girl carrying a bag in one of her hands and another girl carrying a bag in each of her two hands?

Answer:

i) a) When she carries a bag in one hand her centre of mass will shift to the side of the hand that carries the bag.

b) When a bag is in one hand some unbalanced force will act on her and it is difficult to carry.

ii) If she carries two bags in two hands then her centre of mass remains unchanged. Force on two hands are equal i.e. balanced so it is easy to carry the bags.

Question 3.

Two rigid bodies have same moment of inertia about their axes of symmetry. Of the two, which body will have greater kinetic energy?

Answer:

Relation between angular momentum and kinetic energy is, KE = \(\frac{L^2}{2I}\)

Because moment of inertia is same the body with large angular momentum will have larger kinetic energy.

Question 4.

Why are spokes provided in a bicycle wheel? [AP May ’14]

Answer:

The spokes of cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater is the opposition to any change in uniform rotational motion. As a result the cycle runs smoother and speeder. If the cycle wheel had no spokes, the cycle would be driven in jerks and hence unsafe.

Question 5.

We cannot open or close the door by applying force at the hinges. Why? [AP May ’16]

Answer:

To open or close a door, we apply a force normal to the door. If the force is applied at the hinges the perpendicular distance of force is zero. Hence, there will be no turning effect however large force is applied.

Question 6.

Why do we prefer a spanner of longer arm as compared to the spanner of shorter arm?

Answer:

The turning effect of force, τ = \(\overline{\mathrm{r}}\times\overline{\mathrm{F}}\). When arm of the spanner is long, r is larger. Therefore smaller force (F) will produce the same turning effect. Hence, the spanner of longer arm is preferred as compared to the spanner of shorter arm.

Question 7.

By spinning eggs on a table top, how will you distinguish an hard boiled egg from egg? [AP Mar. ’13]

Answer:

To distinguish between a hard boiled egg and a raw egg, we spin each on a table top. The egg which spins at a slower rate shall be a raw egg. This is because in a raw egg, liquid matter inside tries to get away from the axis of rotation. Therefore, its moment of inertia ‘I’ increases. As τ = Iα = constant, therefore, α decreases i.e., raw egg will spin with smaller angular acceleration.

Question 8.

Why should a helicopter necessarily have two propellers?

Answer:

If the helicopter had only one propeller, then due to conservation of angular momen¬tum, the helicopter itself would turn in the opposite direction. Hence, the helicopter should necessarily have two propellers.

Question 9.

If the polar ice caps of the earth were to melt, what would the effect of the length of the day be?

Answer:

Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concen¬trated near the axis of rotation spreads out. Therefore, moment of inertia ‘I’ increases.

As no external torque acts, L = Iω = I(\(\frac{2 \pi}{T}\)) = constant

with increase of I, T will increase i.e., length of the day will increase.

Question 10.

Why is it easier to balance a bicycle in motion?

Answer:

A bicycle in motion is in rotational equilibrium. From principles of Dynamics of rotational bodies is that the forces that are perpendicular to the axis of rotation will try to turn the axis of rotation but necessary forces will arise it cancel these forces due to inertia of rotation and fixed position of axis is maintained. So it is easy to balance a rotating body.

Short Answer Questions

Question 1.

Distinguish between centre of mass and centre of gravity. [AP Mar. 18, 17, 16, 15, 14, 13, May 17; June 15 : TS Mar. 16. 15, May 18, 17]

Answer:

Centre of mass | Centre of gravity |

1) A point inside a body at which the whole mass is supposed to be concentrated. A force applied at this point produces translatory motion. |
1) A point inside a body through which the weight of the body acts. |

2) It pertains (or) contain to mass of the body. | 2) It refers to weight acting on all particles of the body. |

3) In case of small bodies centre of mass and centre of gravity coincide. (Uniform gravitational field) | 3) In case of a huge body centre of mass and centre of gravity may not coincide. (Non uniform gravitational field) |

4) Algebraic sum of moments of masses about centre of mass is zero. | 4) Algebraic sum of moments of weights about centre of gravity is zero. |

5) Centre of mass is used to study translatory motion of a body when it is in complicated motion. | 5) Centre of gravity is used to know the stability of the body where it is to be supported. |

Question 2.

Show that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass. [AP May ’18]

Answer:

Consider a system of particles of masses m_{1}, m_{2}, ……….. mn moves with velocity

But Force (F) = ma, so total force on the body is

F = Ma_{C.M} = m_{1}a_{1} + m_{2}a_{2} + m_{3}a_{3} + ……….. + m_{n} a_{n}

or Total Force F = Ma_{C.M} = F_{1} + F_{2} + F_{3} + ……… + F_{n}

Hence, total force on the body is the sum of forces on individual particles and it is equals to force on centre of mass of the body.

Question 3.

Explain about the centre of mass of earth-moon system and its rotation around the sun.

Answer:

The interaction of earth and moon does not effect the motion of centre of mass of earth and moon system around the sun. The gravitational force between earth and moon is internal force. Internal forces cannot change the position of centre of mass.

The external force acting on the centre of mass of earth and moon system is force between the sun and C.M. of earth, moon system. Motion of centre of mass depends on external force. Hence, earth moon system continues to move in an elliptical path around the sun. It is irrespective of rotation of moon around earth.

Question 4.

Define vector product. Explain the properties of a vector product with two examples. [AP Mar. ’17, ’15 ; TS Mar. ’17, ’16, ’15; APMay ’18. ’17; TS May ’18. ’16]

Answer:

Vector product (or) cross product :

If the product of two vectors (say \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\)) gives a vector then that multiplication of vectors is called cross product or vector product of vectors.

Properties of cross product:

1. Cross product is not commutative i.e.

2. Cross product obeys distributive law i.e.,

3. If any vector is represented by the combination of \(\overline{\mathrm{i}},\overline{\mathrm{j}}\) and \(\overline{\mathrm{k}}\) then cross product will obey right hand screw rule.

4. The product of two coplanar perpendicular unit vectors will generate a unit vector perpendicular to that plane

5. Cross product of parallel vectors is zero

Examples of cross product:

1) Torque (or) moment of force (\(\overline{\mathrm{\tau}}\)) :

It is defined as the product of force and perpendicular distance from the point of application.

∴ Torque τ = \(\overline{\mathrm{r}}\times\overline{\mathrm{F}}\)

2) Angular momentum and angular velocity :

For a rigid body in motion, Angular momentum (\(\overline{\mathrm{L}}\)) = radius (\(\overline{\mathrm{r}}\)) x momentum (\(\overline{\mathrm{P}}\))

∴ Angular momentum (\(\overline{\mathrm{L}}\)) = \(\overline{\mathrm{r}}\) × (m\(\overline{\mathrm{v}}\)) = m(\(\overline{\mathrm{r}}\times\overline{\mathrm{v}}\))

Question 5.

Define angular velocity. Derive v = r ω. [TS Mar. 19,’ 17, 16; AP Mar. 19, May. 16; May 14]

Answer:

Angular velocity (ω) :

Rate of change of angular displacement is called angular velocity.

Relation between linear velocity (v) and angular velocity (ω) :

Let a particle P is moving along circumference of a circle of radius r1 with uniform speed v. Let it is initially at the position A, during a small time ∆t it goes to a new position say C from B. Angle subtended during this small interval is say dθ.

By definition angular velocity,

Question 6.

State the principle of conservation of angular momentum. Give two examples.

Answer:

Law of conservation of angular momentum:

When no external torque is acting on a body then the angular momentum of that rota-ting body is constant.

i.e., I_{1}ω_{1} = I_{2}ω_{2} (when τ = 0)

Example -1:

A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω_{1} (n_{1}revolutions/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I_{1}. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I_{2} and let ω_{2} is his angular speed.

Here ω_{2} < ω_{1} because moment of inertia increases.

Example – 2 :

An athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

Question 7.

Define angular acceleration and torque. Establish the relation between angular acceleration and torque. [TS Mar. ’18, ’17; TS May ’17, June ’15; AP Mar. ’19, June ’15]

Answer:

Angular acceleration (α) :

Rate of change of angular velocity is called angular acceleration.

Torque :

It is defined as the product of the force and the perpendicular distance of the point of application of the force from that point.

Relation between angular acceleration and Torque:

We know that, L = Iω

On differentiating the above expression with respect to time ‘t’

But \(\frac{dL}{dt}\) is the rate of change of angular momentum called ‘Torque (τ)”.

and \(\frac{d \omega}{dt}\) is the rate of change of angular velocity called “angular acceleration (α)”

∴ The relation between Torque and angular acceleration is, τ = lα

Question 8.

Write the equations of motion for a particle rotating about a fixed axis.

Answer:

Equations of rotational kinematics :

If ‘θ’ is the angular displacement, Wj is the initial angular velocity, ω_{f} is the final angular velocity after a time ‘t’ seconds and ‘α’ is the angular acceleration, then the equations of rotational kinematics can be written as,

Question 9.

Derive expressions for the final velocity and total energy of a body rolling without slipping.

Answer:

A rolling body has both translational kinetic energy and rotational kinetic energy. So the total K.E energy of a rolling body is,

Long Answer Questions

Question 1.

(a) State and prove parallel axis theorem.

(b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown into two equal pieces, then find the radius of gyration of each piece about AB.

Answer:

a) Parallel axis theorem :

The moment of inertia of a rigid body about an axis passing through a point is the sum of moment of inertia about parallel axis passing through centre of mass (I_{G}) and mass of the body multiplied by Square of distance (MR²) between the axes i.e.,

I = I_{G} + MR²

Proof :

Consider a rigid body of mass M with ‘G’ as its centre of mass. Iq the moment of inertia about an axis passing through centre of mass. I = The moment of inertia about an axis passing through the point ’O’ in that plane.

Let perpendicular distance between the axes is OG = R (say)

Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from ’P’ on to it as shown in figure.

The moment of inertia about the axis passing through centre of mass G.

(I_{G}) = ∑mGP² ……….. (1)

M.O.I. of the body about an axis passing through ‘O’ (I) = ∑mOP² ………… (2)

From triangle OPD

OP² = OD² + DP²

⇒ OD = OG + GD

∴ OD² = (OG + GD)² = OG² + GD² + 2OG. GD ………….. (3)

From Equations (2) and (3)

I = ∑mOP² = Em [ (OG² + GD² + 2OG. GD) + DP²]

∴ I = ∑m {GD² + DP² + OG² + 20G. GN}

But GD² + DP² = GP²

∴ I = ∑m {GP² + OG² + 20G. GD}

∴ I = ∑m GP² + ∑mOG² + 20G ∑mGD ………. (4)

But the terms ∑mGP² = I_{G}

∑mOG² = MR² (∵ ∑m = M and OG = R)

The term 20G ∑mGD = 0. Because it represents sum of moment of masses about centre of mass. Hence its value is zero.

∴ I = I_{G} + MR²

Hence parallel axis theorem is proved.

b) Moment of inertia of a disc of mass ‘M’ and radius ‘R’ about its diameter is,

I = \(\frac{MR^2}{4}\)

If ‘k’ is radius of gyration of disc then, I = Mk²

∴ Mk² = \(\frac{MR^2}{4}\) ⇒ k = R/2

After cutting along the diameter, mass M of each piece = \(\frac{M}{2}\)

Moment of inertia of each piece,

Question 2.

(a) State and prove perpendicular axis theorem.

(b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axis. Then find the ratio of their radii.

Answer:

a) Perpendicular axis theorem :

The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moment of inertia about two perpendicular axis concurrent with perpendicular axis and lying in the plane of the body.

∴ I_{z} = I_{x} + I_{y}

Proof :

Consider a rectangular plane lamina. X and Y are two mutually perpendicular axis in the plane. Choose another perpendicular axis Z passing through the point ‘O’.

Consider a particle P in XOY plane.

Its co-ordinates are (x, y).

Moment of inertia of particle about

X-axis is I_{X} = ∑my².

M.O.I about Y-axis is I_{Y} = ∑mx²

M.O.I about Z axis is I_{Z} = ∑ m . OP²

From triangle OAP,

OP² = OA² + AP² = y² + x²

∴ Iz = ∑ mOP² = ∑ m (y² + x²)

∴ Iz = X my² + ∑ mx²

But ∑ my² = I_{x} and ∑ mx² = I_{y}

∴ Moment of Inertia about a perpendicular axis passing through ‘O’ is I_{Z} = I_{X} + I_{Y}

Hence perpendicular axis theorem is proved.

b) Moment of inertia of a thin circular ring about its diameter is, I_{1} = m_{1} R²_{1}

Moment of inertia of a flat circular disc about its diameter is, I_{2} = \(\frac{m_2R^{2}_{2}}{2}\)

Given that two objects having same moment of inertia i.e., I_{1} = I_{2}

Question 3.

State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [AP Mar. ’16]

Answer:

Law of conservation of angular momentum: When no external torque is acting on a body then the angular momentum of that rotating body is constant.

i.e., I_{1}ω_{1} = I_{2}ω_{2} (when τ = 0)

Explanation:

Here I_{1} and I_{2} are moment of inertia of rotating bodies and ω_{1} and ω_{2} are their initial and final angular velocities. If

Example -1 :

A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω_{1} (n_{1} revolutions/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I_{1}. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I_{2} and let ω_{2} is his angular speed.

Here ω_{2} < ω_{1} because moment of inertia increases.

Example – 2 :

An athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

Position of centre of mass of some symmetrical bodies :

Shape of the body | Position of centre of mass |

1. Hollow sphere (or) solid sphere | At the centre of sphere |

2. Circular ring | At the centre of the ring |

3. Circular disc | At the centre of disc |

4. Triangular plate | At the centroid |

5. Square plate | At the point of intersection of diagonals |

6. Rectangular plate | At the point of intersection of diagonals |

7. Cone | At \(\frac{3h}{4}\) th of its height from its apex on its own axis |

8. Cylinder | At the midpoint of its own axis. |

Comparison of translatory and rotatory motions :

Problems

Question 1.

Show that a • (b × c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors a, b and c. (IMP)

Solution:

Let a parallelopiped be formed on three

Now \(\hat{a}\) • (\(\hat{b}\times\hat{c}\) x c) = \(\hat{a}\) • be \(\hat{n}\) = (a) (be) cos 0° – abc

Which is equal in magnitude to the volume of parallelopiped.

Question 2.

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope ? Assume that there is no slipping.

Soution:

Here M = 3 kg ; R = 40 cm = 0.4 m

Moment of inertia of the hollow cylinder about its axis, I = MR² = 3(0.4)² = 0.48 kg m²

Force applied, F = 30 N

∴ Torque, τ = F × R = 30 × 0.4 = 12 N – m

If α is angular acceleration produced, then from τ = Iα

Linear acceleration, a = Ra = 0.4 × 25

= 10 ms^{-2}.

Question 3.

A coin is kept a distance of 10 cm from the centre of a circular turn table. If the coefficient of static friction between the table and the coin is 0.8. Find the frequency or rotation of the disc at which the coin will just begin to slip.

Solution:

Distance of coin = r = 10 cm = 0.1 m.

Coefficient of friction µ = 0.8.

Frequency of rotation = number of rotations per second.

Question 4.

Find the torque of a force \(\mathbf{7} \overrightarrow{\mathbf{i}}+\mathbf{3} \overrightarrow{\mathbf{j}}-5 \overrightarrow{\mathbf{k}}\) about the origin. The force acts on a particle whose position vector is \(\overrightarrow{\mathbf{i}}-\overrightarrow{\mathbf{j}}+\overrightarrow{\mathbf{k}}\). [AP Mar. ’14, ’13; May ’13]

Answer:

Question 5.

Particles of masses 1g, 2g, 3g….., 100g are kept at the marks 1 cm, 2 cm, 3 cm…, 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.

Solution:

Given Masses of 1g, 2g, 3g 100 g are 1, 2, 3 ……….. 100 cm on a scale.

i) Sum of masses 2m = \(\sum_{i=1}^n\)ni

Sum of n natural numbers S

∴ Total mass M = 5051 gr = 5.051 kg → (1)

ii) Centre of mass of all these masses is given by

iii) M.O.I. = I

Sum of cubes of 1st n natural numbers is

M.O.I. about C.M. = I_{G} = I – MR²

= 2.550 – 5.05 × 0.67 × 0.67 = 2.550 – 2.267 = 0.283 kg.m2

iv) Perpendicular bisector is at 50 CM.

So shift M.O.I from centre of mass to

x_{1} = 50cm point from x = 67 CM

∴ Distance between the axis

R = 67 – 50 = 17cm = 0.17M

M.O.I. about this axis I = I_{G} + MR²

= 0.283 + 5.05 × 0.17 × 0.17

= 0.283 + 0.146 = 0.429 kgm²

∴ M.O.I. about perpendicular bisector of scale = 0.429 kg – m²

Question 6.

Calculate the moment of inertia of a fly wheel, if its angular velocity is increased from 60 r.p.m. to 180 r.p.m. when 100 J of work is done on it. [TS May ’16]

Solution:

W = 100 J, ω_{1} = 60 RPM = 1 R.P.S = 2π Rad.

ω_{2} = 180 R.P.M. = 3 R.P.S = 6π Rad.

Question 7.

Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.

Solution:

Mass of each particle m = 100 g; side of equilateral triangle = 10 cm.

In equilateral triangle height of angular bisector CD = \(\frac{\sqrt{3}}{2}\)l

Centroid will divide the angular bisector in a ratio 2 : 1

So X distance of each mass from vertex to centroid is 2.\(\frac{\sqrt{3}}{2}\)l = \(\frac{\sqrt{3}}{2}\)l

Moment of Inertia of the system

Question 8.

Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.

Solution:

Mass of each particle, m = 100 g = 0.1 kg.

Length of side’of square, a = 10 cm = 0.1 m

In square distance of corner from centre of square = \(\frac{1}{2}\) diagonal = \(\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}\)

∴ Total moment of Inertia

Question 9.

Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.

Solution:

Mass of disc = M = 1 kg.

Radius of disc = 20 cm = 0.2 m

They are in contact as shown.

M.O.I of a circular disc about a tangent parallel to its plane = \(\frac{5}{4}\) MR²

Total M.O.I. of the system

Question 10.

Four spheres each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.

Solution:

Diameter of sphere = 2a ⇒ radius = a.

Side of square = b.

For spheres 1 and 2 axis of rotation is same and passing through diameters. M.O.I. of solid sphere about any diameter = \(\frac{2}{5}\)MR² (put M = m and R = a)

Transfer this M.O.I. on to the axis using

Parallel axis theorem.

Total M.O.I. of the system

Question 11.

To maintain a rotor at a uniform angular speed or 200 rad s^{-1}, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

Solution:

Given uniform angular speed (ω) = 200 rad s^{-1}

Torque, τ = 180 N – m ; But power p = τω

∴ P =180 × 200 = 36000 watt = 36 kW

Question 12.

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Solution:

Let m be the mass of the stick concentrated at C, the 50 cm mark

For equilibrium

about C’, i.e. at the 45 cm mark,

10g (45 – 12) = mg (50 – 45)

10g × 33 = mg × 5

m = \(\frac{10\times33}{5}\) = 66 grams

Question 13.

Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5 kg and radius 1 m.

Solution:

Mass of disc, M = 5 kg; Radius R = 1 m.

Angular velocity, co = 60 RPM = \(\frac{60\times 2\pi}{60}\) = 2πRad/sec

M.O.I. of disc about a point passing through circumference and perpendicular to the plane.

Question 14.

Two particles, each of mass m and speed u, travel in opposite directions along para¬llel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momemtum is taken.

Solution:

Angular momentum, L = mvr

Choose any axis say ‘A’

Let at any given time distance between m_{1} & m_{2} = L = L_{1} + L_{2}

About the axis ‘A’ both will rotate in same direction See fig.

∴ Total angular momentum

L = L_{1} + L_{2} = muL_{1} + muL_{2} = mu (L_{1} + L_{2}) = muL

about any new axis say B distance of m_{1} and m_{2}

are say L’_{1} and L’_{2}

Total angular momentum,

L = mu L’_{1} + mu L’_{2}

or L = mu(L’_{1} + L’_{2}) = muL (∵ L’_{1} + L’_{2} = L)

Hence, total angular momentum of the system is always constant.

Question 15.

The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm². Find the torque required to bring it to rest in 20s.

Solution:

M.O.I, I = 0.3 kg. ; time, t = 20 sec.,

ω_{1} = 300 R.P.M. = \(\frac{300}{60}\) = 5. R.P.S.; ω_{2} = 0

Torque, τ = Iα = 0.3 × \(\frac{5\times 2\pi}{20}\) = 0.471 N – m.

Question 16.

When 100J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel?

Solution:

W=100J, ω_{1} = 60 RPM = 1 R.PS = 2π Rad.

ω_{2} =180 R.P.M. = 3 R.P.S = 6π Rad.

Question 17.

Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are lOOg, 150g and 200g respectively. Each side of the equilateral triangle is 0.5 m long, lOOg mass is at origin and 150g mass is on the X-axis. [TS Mar. 18, June 15; AP Mar. ’18]

Solution:

Mass at A = 100g ; Coordinates = 0, 0

Mass at B = 150 g; Coordinates = (0.5, 0)

Mass at C = 200g; Coordinates (0.25,0.25 √3 )

Coordinates x_{cm}