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**System of Particles and Rotational Motion**

**System of Particles and Rotational Motion****Chapter: 6**

Part – I |

**EXERCISE**

**1. Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?**

Ans: (i) The centre of mass of the sphere is located at its centre.

(ii) The centre of mass of a cylinder is at the centre of its axis of symmetry.

(iii) The centre of mass of a ring is at the centre of the ring.

(iv) The centre of mass of a cube is at its geometrical centre.

Hence, the centre of mass of these bodies will lie in their respective geometric centers. The centre of mass of a body may or may not lie within it, depending on their geometry. As in the case of a hollow sphere the centre of mass lies outside the body while for a solid sphere, centre of mass lies within its body.

**2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.**

Ans: x cm = m1×1 + m2×2/ m1 + m2

Here m1 and m2 are the masses of the hydrogen (H) and chlorine (CI) atoms, respectively and x1 and x2 are their positions:

Let mH = m

Then mCI = 35.5m

**The centre of mass is given by:**

**The mH terms cancel out:**

xCM = 35.5 × 1.27 / 36.5

xCM = 45.085 / 36.5

xCM = 1.235 Å.

**3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?**

Ans: 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. Since there are no external horizontal forces acting on the system, the total horizontal momentum of the system remains constant. As a result, the child and trolley have a centre of mass with constant speed.

**4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.**

Ans:

**5. Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c.**

Ans: A parallelopiped with origin O and sides a, b and c is shown in the following figure.

Volume of parallelopiped = abc.

6. Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px , py and pz . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Ans:

L_{y} = zp_{x} – xp_{z}

L_{z} = xp_{y} – yp_{x}

Where Lx, Ly and Lz are the components of angular monumentum L in x, y, z plane, then z = 0 and p_{z} = 0

Hence, from (i) we have

L_{x} = 0, L_{y} = 0

L_{z} = xp_{v} – yp_{x}

Thus, angular momentum has only z – component if the particle moves only in the X – Y plane.

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

Ans:

Vector angular momentum of the system about any point P on x_{1}y1

Similarly vector angular momentum about a point Q on x2y2

The vector angular momentum about R,

Thus, the vector angular momentum of the given two particles will be the same irrespective of the point taken.

**8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.**

Ans: Here, α = 36.9°, β = 53.1°

If T_{1}, T_{2} are tensions in the string. Then for equilibrium along the horizontal

T_{1}, sin α = T_{2} sin β

T_{1}/T_{2} = sin β / sin α = sin 53.1°/ sin 36.9°

= 0.747 / 0. 5477

= 1.3523.

d is the distance of the centre of gravity G of the bar from the left end.

**From the horizontal equilibrium, we have:**

T_{1} × sin α = T_{2} × sin β

= T_{1}/T_{2} = sin β / sin α = 1.3523

**Now, for rotational equilibrium:**

T_{1} × cos α × d = T_{2} × cos β × (2 – d)

Substituting T_{1} = 1.3523 × T_{2}:

1.3523 × T_{2} × cos α × d = T_{2} × cos β × (2 – d)

**Cancelling T**_{2}**:**

1.3523 × cos α × d = cos β × (2 – d)

**Substituting the values of α and β:**

1.3523 × cos 36.9° × d = cos 53.1° × (2 – d)

**Solving for d:**

d = 1.15 m.

9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Ans: Weight of the car: W = 1800 kg × 9.8 m/s^{2} = 17658 N

Distance between front and back axles: L = 1.8 m

Distance of the centre of gravity from the front axle: d = 1.05m

**Sum of vertical forces:**

F_{f} + F_{b} = W

F_{f }+ F_{b} = 17658 N

**Sum of moments about the Front Axle:**

F_{b} . L = W . d

F_{b} . 1.8 = 17658 . 1.05

F_{b} . 18 = 18541.9

F_{b }= 18541.9 / 1.8

= 10245.5 N

Find F_{f}:

F_{f} + F_{b} = 17658

F_{f} + 10245.5 = 17658

F_{f }= 17658 – 10245.5

= 7402.5 N.

**10. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. **

Ans: **Moment of inertia:**

*I* _{cylinder} = mr^{2}

Here m is the mass and r is radius.

*I *_{sphere} = ⅖ mr^{2}

**Relationship between torque and Angular Acceleration:**

T = *I*α

Α = T/* I*

**Angular Acceleration for each object:**

Α_{cylinder} = T/ *I*_{cylinder} = T/mr^{2}

**For the solid sphere:**

**Comparison of Angular Accelerations:**

αsphere = 5/2 α^{cylinder}

**Angular speed after a given time:**

**The angular speed (ω) after a given time (t) is related to the angular acceleration (α) by:**

ω = αt

For the hollow cylinder

ω_{cylinder} = α_{cylinder}t = (T/mr^{2}) t

**For the solid sphere:**

ω_{ sphere} = α_{sphere}t = (5r/ 2mr^{2})t

**Comparison of Angular Speeds:**

**We know that:**

ω = αt

α_{sphere} = (5/2) × α_{cylinder}

Substituting the value of αsphere in the equation for

ω_{sphere:}

ω_{sphere} = ((5/2) x α_{cylinder}) × t

**Comparison of Angular Speeds:**

ω_{cylinder} = α_{cylinder} × t

ω_{sphere} = (5/2) × α_{cylinder} × t

Therefore, ω_{sphere} = (5/2) × ω_{cylinder}.

**11. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?**

Ans: **Angular Momentum:**

**Given data:**

Mass (m) = 20 kg

Angular speed (ω) = 100 rad/s

Radius (r) = 0.25m

**Calculate the moment of inertia (***I***):**

= ½ mr^{2} = ½ × 20kg × (0.25)^{2}

= ½ x 20 × 0.0625

= 0.625 kg.

**Calculate the moment of inertia (k):**

½ I ω^{2} = ½ × 0.625 kg . m^{2} × (100 rad/s)^{2 }

= ½ x 0.625 × 10000

**Calculate the angular momentum (L):**

L = *I*ω = 0. 625 kg . m^{2} × 100 rad/s = 62.5 kg .m^{2}/s.

12. (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

Ans: (a) I_{2} = ⅖ I1,

v_{1} = 40 rev . min^{-1}

By using the principle of conservation of angular momentum, we get

I_{1}ω_{1} = I_{2} ω_{2}

ω_{2} = I_{1}ω_{1} / I_{2}

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Ans: Final K.E. of rotation / Initial K.E. of rotation

= 2.5

Or (K.E.) final = 2.5 (K.E.) Initial.

Clearly, final (K.E.) rot becomes more, because the child uses his internal energy when he folds his hands to increase the kinetic energy.

**13. A rope of negligible mass is wound round a hollow cylinder of 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. **

Ans: Here, M = 3kg,

R = 40 cm = 0.4 cm.

M.I. of the hollow cylinder about its axis

= I = MR2

= 3 × (0.4)2 = 0.48 kg m2

When the force of 30 N is applied over the rope wound round the cylinder, the torque will act on the cylinder. It is given by,

= 30 × 0.4 = 12 Nm.

If α be the angular acceleration produced, then,

or, α =

= 12/ 0.48

= 25 rad. s-2.

14. To maintain a rotor at a uniform angular speed of 200 rad s^{-1}, an engine needs to transmit a torque of 180 N m. 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.

Ans: **Given that:**

Angular speed (ω) = 200 rad/s

Torque (T) = 180 Nm

**Power (P):**

P = T × ω

P = 180 Nm × 200 rad/s

= 36000W or 36kW.

15. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Ans: Assuming mass per unit area of disc = m mass of the original disc

= M = 𝝅R^{2} × m

Mass of the portion removed

= M’ = 𝝅R2 / 4

m = M/4

As shown the mass M is concentrated at O and the mass M’ is concentrated at O’ and OO’ = R/2

**If x is the distance of the centre of mass of the remaining part, then:**

Negative values shows that P os to the left of O.

**16. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g 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? **

Ans: Let W and W be the respective weights of the metre stick and the coin. The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.

**Calculate the distance:**

dc = 45 cm – 12 cm = 33 cm

Distance of the centre of the metre stick from the new balance point is 45 cm.

dm = 50 cm – 45 cm = 5 cm.

**Set up the torque balance equation:**

Clockwise torque = Counterclockwise torque

m × g × dm = (2mc) × g × dc

m × 5 cm = 10g × 33 cm

m × 5 = 10 × 33

m × 5 = 330

m = 330/5

66 g.

**17. The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 × 10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.**

Ans: **Given that:**

m = 5.30 × 10^{-26} kg

*I *= 1.94 × 10^{-46} kg × m^{2}

v = 500 m/s

E_{rot} = ⅔ E_{trans}

**Calculate the translational kinetic E**_{trans}**:**

E_{trans} = ½ mv^{2}

E_{trans} = ½ × 5.30 × 10^{-26} kg x (500m/s)^{2}

E_{trans} = ½ × 5.30 × 10^{-26} × 250000

E_{trans} = ½ × 1.325 × 10^{-21}

E_{trans} = ½ 6 . 625 × 10^{-22} J

**Calculate the rotational Kinetic energy E**_{rot}**:**

E_{rot} = ⅔ E_{trans}

E_{rot} = ⅔ × 6.625 × 10^{-22} J

E_{rot} = 4.417 × 10^{-22} J

**The rotational kinetic energy is also given by:**

E_{rot} = ½ *I *ω^{2}

ω^{2} = 2Erot / *I*

ω^{2 }= 4.5546 × 10^{24}

ω = √4.5546 × 10^{24}

ω = 6.75 × 10^{12} rad/s.