Q:1. Give the location of the centre of mass of a (i)Sphere (ii)cylinder (iii)ring (iv)cube , each of uniform mass density . Does the centre of mass of a body necessary lie on the body ?
Sol: As mass density is uniform in all the cases , hence centre of mass is located at their respective geometrical centre .
It is not necessary that the centre of mass of a body should lie on the body . e.g. in case of circular ring , centre of mass is at the centre of the ring , where there is no mass .
Q: 2.In the HCI molecule, the separation between the nuclei of two atoms is about 1.27 Å (1Å = 10-10m). Find the approximate location of the C.M. 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?
Sol. Let m → mass of H – atom
35.5 m → mass of CI – atom
The C.M. is situated at a distance x from H-atom.
Distance of C.M. from CI atom = (1.27 – x) Å
Considering the C.M. as the origin, the sum of moments of both atoms about the origin will be zero.
$\displaystyle 0 = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} $
$\displaystyle m_1 x_1 + m_2 x_2 = 0 $
mx – (1.27 – x) 35.5m = 0 [both have opposite direction]
x = 35.5 (1.27 – x).
x + 35.5x = 35.5 × 1.27
36.5x = 35.5 × 1.27
x = (35.5 × 1.27)/36.5
= 1.235 Å
Q: 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 C.M. of the system?
Sol. When the child gets up and runs on the trolley, the forces acting on the system (child + trolley) are internal. Since, no external force is acting on the system, the speed of the C.M. of the system remains unchanged.
Q:4. Show that the area of the triangle contained between the vectors $\vec{a}$ and $\vec{b}$ is one half of the magnitude of $\vec{a}\times \vec{b}$
Sol. Let $ \vec{OA}=\vec{a}$ and $ \vec{OB}=\vec{a}$ and ∠AOB = θ . complete the parallelogram OACB. Join BA. Draw BN ⟘ OA.
In ∆ OBN,
Sin θ = BN/OB = BN/b
BN = bsinθ
$|\vec{a}\times \vec{b}|= a b sin\theta = OA\times BN$
$ = \frac{2(OA \times BN)}{2}$
= 2 (area of ∆OAB)
Area of ∆ OAB $= \frac{1}{2}|\vec{a}\times \vec{b}|$
Q:5. Show that $ \vec{a}.(\vec{b}\times \vec{c})$ is equal in magnitude to the volume of the parallelopied formed on the three vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$
Sol.
$\vec{b}\times \vec{c} = b c sin90 \hat{n}$
$= b c \hat{n}$ where $\hat{n}$ is the unit vector along $\vec{OA}$ perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ .
$\vec{a}.(\vec{b}\times \vec{c}) = (\vec{a}).bc\hat{n}$
= (a) (bc) cos 0 (∵ $\hat{n}$|| $\vec{a}$ )
= abc = volume of the parallelepiped.
Q:6. Find the components along the X , Y , Z axes of the angular momentum $\vec{L}$ of a particle , whose position vector is $\vec{r}$ with components x , y , z and momentum is $\vec{p}$ 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 .
Q:7. Two particles each of mass m speed v , travel in opposite direction along parallel 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 momentum is taken .
Sol: 
Angular momentum of two particle system about any point A on X1 Y1
$\large L_A = m v \times 0 + m v d = m v d $
Angular momentum of two particle system about any point B on X2 Y2
$\large L_B = m v \times d + m v \times 0 = m v d $
Let us consider any point C on AB , where A C = x
Angular momentum of two particle system about point C on is
$\large L_C = m v \times x + m v \times (d-x) = m v d $
$\large L_A = L_B = L_C $
Q: 8.A non – uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angels made by the strings with the vertical are 36.90 and 53.10 respectively. The bar is 2m long. Calculate the distance ‘d’ of the centre of gravity of the bar from its left end.
Sol. Let θ1 = 36.9° and θ2 = 53.1°
T1 and T2 are tensions in the two strings. For equilibrium for the rod along the horizontal,
T1 sinθ1 = T2 sinθ2
$ \displaystyle \frac{T_1}{T_2} = \frac{sin\theta_2}{sin\theta_1}$
$\displaystyle \frac{T_1}{T_2} = \frac{sin 53.1}{sin 36.9}$
$\displaystyle \frac{T_1}{T_2} = \frac{0.7407}{0.5477}$
$ \displaystyle \frac{T_1}{T_2} = 1.3523 $
Let C be the position of centre of gravity of the rod from the left and at a distance d.
For rotational equilibrium of the rod about C, the moment of the vertical forces must be equal and opposite.
T1 cos Ɵ1 × d = T2 cos Ɵ2 (2 – d)
T1 cos 36.90 × d = T2 cos 53.10 (2 – d)
$ \displaystyle \frac{T_1}{T_2} \times \frac{cos36.9}{cos53.1} = \frac{2-d}{d} $
$ \displaystyle \frac{T_1}{T_2} \times \frac{0.8366}{0.6718} = \frac{2-d}{d} $
$ \displaystyle 1.3523 \times \frac{0.8366}{0.6718} = \frac{2-d}{d} $
d = 0.745 m
Q: 9.A car weight 1800 kg. The distance between its front and back is 1.8 m. it’s 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.
Sol. m = 1800 kg,
Let R1, R2 = Reaction forces due to ground on each front wheel each back wheel
R1 + R2 = mg = 1800 × 9.8 … (i)
Distance between the front and back axles = 1.8m
Distance of centre of gravity from the front axle = 1.05 m
For rotational equilibrium of the car about the centre of gravity,
R1 × 1.05 = R2 (1.8 – 1.05) = R2 × 0.75
$ \displaystyle \frac{R_1}{R_2} = \frac{0.75}{1.05} = \frac{5}{7}$ ….(ii)
From (i) and (ii),
$ \displaystyle \frac{5}{7}R_2 + R_2 = 1800 \times 9.8 $
$ \displaystyle R_2 = \frac{1800\times 9.8\times 7}{12} $
R2 = 10290 N
From (i)
R1 = 7350 N
Q: 10. (a) Find the M.I. of sphere about a tangent to the sphere, given the M.I. of the sphere about any of its
diameters to be (2/5) MR2, where M is the mass and R is the radius of the sphere.
(b)What is the moment of inertia of a uniform circular disc of radius R and mass M about an axis
(i) passing through its centre and normal to the disc,
(ii) passing through a point on its edge and normal to the disc? The moment of inertia of the disc about any to its diameter is given to be (1/4)MR2 .
Sol. Given, IAB = (2/5)MR2 , ICD = ?
By parallel axes theorem, ICD = IAB + MR2
(2/5) MR2 + MR2 = (7/5) MR2
Consider 2 diameters AB and CD symmetrical to each other IAB = ICD = (1/4)MR2
(i)By ⟘ axes theorem, M.I. of the disc about an axis passing through its centre and normal to its plane
IXY = IAB + ICD = 2 × (1/4) MR2 = (1/2) MR2
(ii)By parallel axes theorem, M.I. of the disc about an axis passing through its edge and ⟘ to its plane
IPQ = IPY + MR2 = (1/2)MR2 + MR2 = (3/2)MR2