I = \(\frac { 2 }{ 5 } \)MR² where M is the mass and R is radius of the solid sphere.

The moments of inertia will be different on account of different distances.

The radial component of a force does not contribute towards torque.

The stone will move along the tangent at the point of breaking.

No, \(K=\sqrt { \frac { 1 }{ M } } \)

The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig

By symmetry each strip has its centre of mass at its midpoint. If we join the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.

They are of unequal mass. Their masses are in the inverse ratio of the arms of the balance.

In a whirl wind, the air from nearby region gets concentrated in a small space thereby decreasing the value moment of inertia considerably. Since \(I\omega \)= constant, due to decrease in moment of inertia, the angular speed becomes quite high.

Molten ice from poles into ocean and so mass is going away from axis of rotation. So, moment of inertia of earth increases and to conserve angular momentum, angular velocity (\(\omega \)) decreases. So, time period of rotation increases (\(T=2\pi /\omega \) ). So, net effect of global warming is increasing in the duration of day.

The moment of inertia of the system (circular table + two boys) decreases. To conserve angular momentum (\(L=I\omega =constant\)), the speed of rotation of the circular table increases.

All mass of a hollow cylinder at a distance R from axis of rotation. Whereas in case of a sphere, most of mass lies at a distance less than R from axis of rotation. As moment of inertia is \(\sum M_{ i }{ R }_{ i }^{ 2 }\), so sphere as a lower value of moment of inertia.

When the person withdraws his chest, his moment of inertia decreases. No external torque is acting on the system. So, to conserve angular momentum, the angular velocity increases.

Moment of inertia of a sphere is less than that of a cube of same mass.

This is based on the conservation of angular momentum.

10 m/s

Kinetic energy of rotation =  \(\frac { 1 }{ 2 } I{ \omega }^{ 2 }=\frac { 1 }{ 2 } (I\omega )=\frac { 1 }{ 2 } L\omega \)
L= Iω is constant if moment of inertia (I) of the system changes. It means as I changes, then ω also changes to keep Iω= constant. Hence K.E. of rotation also changes with the change in I. In other words, rotation K.E. is not conserved.

According to the law of conservation of angular momentum
I₁ω₁ =I₂ω₂ ⇒ \(\frac { { I }_{ 1 } }{ { T }_{ 1 } } =\frac { { I }_{ 2 } }{ { T }_{ 2 } } \)
or T₂ =\(\frac { { I }_{2 } }{ { I }_{ 1 } } { T }_{ 1 }\)
I₁ =\(\frac { 2 }{ 5 } \)MR²,I₂=\(\frac { 2 }{ 5 } M\left( \frac { R }{ 2 } \right) ^{ 2 }\)
∴  T₁ =\(\frac { 1 }{ 4 } \times 24\)=6 hr

(a)The velocity of centre of mass \(\upsilon \)_cm is given by the expression
\(\upsilon _{ cm }=\frac { { m }_{ 1 }{ \upsilon }_{ 1 }+{ m }_{ 2 }{ \upsilon }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)

Here m₁ = 5 kg, m₂ = 2 kg \(\upsilon _{ 1 }\)=14m/s ,\(\upsilon _{ 2 }\) =0
Substituting these values, we get
\(\upsilon \)_cm =\(\frac { 5\times 14+2\times 0 }{ 5+2 } \) =10m/s
(b)We know that the centre of mass coordinate reference frame is one in which centre of mass is at rest. So the velocity of heavier block in this frame just after the kick is
\(\upsilon '_{ 1 }={ \upsilon }_{ 2 }-{ \upsilon }_{ cm }\)  =14 - 10 = 4 m/s
and that of lighter block is
\(\upsilon '_{ 2 }={ \upsilon }_{ 1 }-{ \upsilon }_{ cm }\) = 0 - 10 = -10 m/s

Torque on cylinder, \(\tau =force\times radius\) 
\(=30\times 40=12N-m\)  
Moment of inertia of hollow cylinder about its axis
\(I={ MR }^{ 2 }=3\times { \left( 0.4 \right) }^{ 2 }=0.48kg-{ m }^{ 2 }\)     
Also,    \(\tau =I\alpha \Rightarrow \alpha \frac { \tau }{ I } \) 
\(\therefore \alpha =\frac { 12 }{ 0.48 } =25{ s }^{ -2 }\)    
Linear acceleration of rope
\(\alpha =\frac { F }{ m } =\frac { 30 }{ 3 } =10m/{ s }^{ 2 }\).              

\(Given,\ { m }_{ 1 }=1kg,{ m }_{ 2 }=2kg\)
\( { x }_{ 1 }=1m,{ x }_{ 2 }=-1m\)
\( { y }_{ 1 }=2m,{ y }_{ 2 }=3m\)
\(\\ \therefore \quad { x }_{ CM }=\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 1+2\times -1 }{ 1+2 } \)
\(=\frac { 1-2 }{ 3 } =\frac { -1 }{ 3 } =-0.33\)
\(\\ and\ { y }_{ CM }=\frac { { m }_{ 1 }{ y }_{ 1 }+{ m }_{ 2 }{ y }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 2+2\times 3 }{ 1+2 } \)
\(\\=\frac { 2+6 }{ 3 } =\frac { 8 }{ 3 } =2.66\)
Thus, the coordinates of the centre of mass are (-0.33, 2.66).

Step 1: Draw diagram showing coordinates of the spheres.


Step 2: Find coordinates of centre of mass.
Formula used: \(x_{C M}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\) and
\(\mathrm{y}_{\mathrm{CM}}=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2}{\mathrm{~m}_1+\mathrm{m}_2}\)
x - coordinates of center of mass
\( x_{c m}=\frac{m \times 0+m \times 2 R+m R}{3 m} \)
\( x_{c m}=R \)
\( y-\text { coordinates of center of mass } \)
\( y_{c m}=\frac{m \times 0+m \times 0+m R}{3 m} \)
\( x_{c m}=\frac{R}{\sqrt{3}}\)
Hence coordinates are, \(\left(R, \frac{R}{\sqrt{3}}\right)\)

Consider an equilateral triangle of side 1 m as shown in fig. Take X and Y axes as shown in fig.
By the definition of centre of mass, we have
\(\vec { x } \) =\(\frac { { m }_{ 1 }{ x }_{ 1 }{ +m }_{ 2 }{ x }_{ 2 }+{ m }_{ 3 }{ x }_{ 3 } }{ { m }_{ 1 }{ +m }_{ 2 }+{ m }_{ 3 } } \)
and  \(\vec { y } =\frac { { m }_{ 1 }y_{ 1 }{ +m }_{ 2 }y_{ 2 }+{ m }_{ 3 }y_{ 3 } }{ { m }_{ 1 }{ +m }_{ 2 }+{ m }_{ 3 } } \)
Here, m₁ = 1 kg, m₂= 2kg and m₃ = 3 kg
[x₁= 0,y₁= 0] [x₂  =1,y₂= 0] and [x₃= 0.5 y₃ = \(\frac { \sqrt { 3 } }{ 2 } \)]
Here y₃ = CD = AC sin 60^o = \(\frac { \sqrt { 3 } }{ 2 } \) 
\(\bar { x } \) = \(\frac { 1\times 0+2\times 1+3\times 0.5 }{ 1+2+3 } =\frac { 3.5 }{ 6 } m\)
and  \(\vec { y } =\frac { 1\times 0+2\times 0+3\times \left( \frac { \sqrt { 3 } }{ 2 } \right) }{ 1+2+3 } =\frac { \sqrt { 3 } }{ 4 } m\)
The co-ordinates of centre of mass are \(\left( \frac { 3.5 }{ 6 } ,\frac { \sqrt { 3 } }{ 4 } \right) \)

Consider a spherical body of mass M and radius r, rolling down on an inclined plane without slipping. Let the angle of inclination of the plane is \(\theta\) with the horizontal (Seefig.)
It \(\omega\) be the angular velocity of the body, then the linear velocity of the body is

\(v=r\omega\)................1
The various forces acting on the body are:
(a) The weight (Mg) of the body in the vertically downward direction
(b) Normal reaction (R) of the surface of the plane on the body which acts vertically upward.
(c) The force of friction (F) which acts opposite to the direction of motion of the body. Resolve Mg into two components:
\(\rightarrow\) Mg cos 8 is the horizontal component, which is equal and opposite to the normal reaction
i.e, R=Mg cos \(\theta\) .............2
\(\rightarrow\) Mg sin 8 is the vertical component, which acts in the direction of the motion of the body
\(\therefore\) Net force acting in the direction of motion of the body = Mg sin\(\theta\) - F
Let a = acceleration of the body
Therefore, equation of motion of the body is given by,
Ma = Mg sin \(\theta\) - F ...........3
The external torque acting on the body is produced by the force of friction (F) and its lever arm is r (the radius of the spherical body) [The lines of action of Mg and R pass through the centre of mass of body and hence do not contribute anything to the torque about the centre of mass].
\(\therefore \tau=Fr\)..........4
If \(\alpha\) be the angular acceleration produced in the body whose moment of inertia is I, then
\(\tau=I\alpha\)............5
From eqns (iv) and (v), we get
\(Fr=I\alpha\ or\ F=I\frac{\alpha}{r}\)................6
Using eqn. (vi) in eqn. (iii), we get
\(\therefore Ma=Mgsin\theta-I\frac{\alpha}{r}\)
\(a=r\alpha\ or \alpha=\frac{a}{r}\)
\(\therefore Ma=Mgsin\theta-I\frac{a}{r^2}\Rightarrow a=gsin\theta-\frac{Ia}{Mr^2}\)
\(a(1+\frac{I}{Mr^2})=gsin\theta\)
\(a=\frac{gsin\theta}{1+(\frac{I}{Mr^2})}\)
\(\alpha=\frac{a}{r}=\frac{gsin\theta}{r[1+\frac{I}{Mr^2}]}\)
Substituting this value in eqn. (vi), we get
\(F=\frac{I}{r}.\frac{gsin\theta}{r[1+\frac{I}{Mr^2}]}=\frac{Igsin\theta}{r^2[1+\frac{I}{Mr^2}]}\)
This is the force of friction, required by the body to roll down an inclined plane without slipping.

(i) a = 6 x 10^-2 m/s²
(ii) \(\alpha \) = 1.20 rad/s²