Pure Rotational Motion: Definition, Characteristics and Terminologies

Pure Rotational Motion: Rotational motion and Translational motion are the two types of motion that we see in our surroundings. The translation is a motion of a body along a straight line, but rotation is the motion of wheels, motors, gears, planets, hands of a clock, the rotor of jet engines, and helicopters’ blades.
Imagine an ice skater gliding across the ice in a straight line with constant speed. Her motion is called translation. Consider another skater spinning at a constant rate about a vertical axis at the same point. This kind of motion is called Rotational motion.

We have seen tornadoes and cyclones, which are so devastating? We can’t say that the speed of rotation of tornados is the only governing factor of the devastation caused by them. A tornado is a mixture of force and energy. These govern the rotational motion of a tornado, resulting in destructions.
This article will deal with the concept of the Centre of mass of a system of particles in rotational motion and terminologies related to rotational motion.

Learn All Concepts on Rolling Motion

Some Important Definitions

(1) Rigid body: A rigid body can rotate with all the parts locked together without changing its shape. The linear distance between any two particles in a rigid body always remains the same.

(2) System: A collection of particles interacting with one another and are analysed during a situation are said to form a system.
(3) Pure rotational motion: Pure Rotational motion is the type of motion about a fixed axis. All the particles constituting it undergoes circular motion about a common axis, then that type of motion is rotational motion.

A rotating body is said to be in pure rotation if all the points at the same radius from the Centre of rotation will have the same velocity.

Centre of Mass (COM)

The Centre of Mass (COM) of a system or body is a point that moves as if all the mass were concentrated or integrated at that point and all external forces were applied there.

The position vector of the centre of mass for n particle system

Let a system consist of \(n\) particles of masses \(m_1,\,m_2,\,m_3,…….m_n\) whose positions vectors are \(\overrightarrow {{r_1}} ,\,\overrightarrow {{r_2}} ,\,\overrightarrow {{r_3}} ……..\overrightarrow {{r_n}} \), respectively.

The position vector of the Centre of mass \(\overrightarrow r = \frac{{{m_1}\overrightarrow {{r_1}} + {m_2}\overrightarrow {{r_2}} + {m_3}\overrightarrow {{r_3}} + …………..{m_n}\overrightarrow {{r_n}} }}{{{m_1} + {m_2} + {m_3} + …………{m_n}}}\)
Hence, the Centre of particles’ mass is a weighted average (arithmetic mean) of the position vectors of \(n\) particles making up the system.

Note:

1. The position vector of COM for the two-particle system \(\overrightarrow r = \frac{{{m_1}\overrightarrow {{r_1}} + {m_2}\overrightarrow {{r_2}} }}{{{m_1} + {m_2}}}\), and the Centre of mass (COM) lies between the particles on the line joining the particles.
2. If two masses are equal, i.e. ,\(m_1 = m_2\) then position vector of Centre of mass \(\overrightarrow r = \frac{{\overrightarrow {{r_1}} + \overrightarrow {{r_2}} }}{2}\)

Important points about centre of mass

1. The position of COM is independent of the coordinate system chosen.
2. The position of COM depends upon the distribution of mass and the shape of the body.
   Example: The COM of a circular disc (flywheel) is within the material of the body, while that of a circular ring (bangle) is outside the material of the body.
3. In symmetrical bodies where mass distribution is homogenous, the COM coincides with the geometrical Centre of the body.
4. The COM changes its position only during the translatory motion, and there is no effect of rotatory motion on the COM of the body.
5. If the origin \((0,\,0)\) is at the Centre of mass (COM), then the sum of the moments (distance \(\times\) mass) of the masses of the system about the Centre of mass (COM) is zero,
   i.e., \(\sum {{m_i}\overrightarrow {{r_i}} } = 0\).
6. Consider a system of particles of masses \({m_1},{m_2},{m_3},……\) that moves with velocities, \({v_1},{v_2},{v_3},……\) then the velocity of Centre of mass is given as \({v_{{\text{cm}}}} = \frac{{\sum {m_i}{v_i}}}{{\sum {m_i}}}\).
7. Consider a system of particles of masses \({m_1},{m_2},{m_3},……\) move with accelerations \({a_1},{a_2},{a_3},……\) then the acceleration of Centre of mass is given by \({A_{\text{cm}}} = \frac{{\sum {m_i}{a_i}}}{{\sum {m_i}}}\)
8. Let \(\overrightarrow r\) is a position vector of COM of a system, the velocity of Centre of mass is given by \({\overrightarrow v _{{\text{cm}}}} = \frac{{d\overrightarrow r }}{{dt}} = \frac{d}{{dt}}\left( {\frac{{{m_1}\overrightarrow {{r_1}} + {m_2}\overrightarrow {{r_2}} + {m_3}\overrightarrow {{r_3}} + ……}}{{{m_1} + {m_2} + {m_3} + ……}}} \right)\)
9. Acceleration of Centre of mass \({\overrightarrow A {{\text{cm}}}} = \frac{{d{{\overrightarrow v }{cm}}}}{{dt}} = \frac{{{d^2}\overrightarrow r }}{{d{t^2}}} = \frac{{{d^2}}}{{d{t^2}}}\left( {\frac{{{m_1}\overrightarrow {{r_1}} + {m_2}\overrightarrow {{r_2}} + …….}}{{{m_1} + {m_2} + {m_3} + …….}}} \right)\)
10. Force on a rigid body \(\overrightarrow F = M{\overrightarrow A _{{\text{cm}}}} = M\frac{{{d^2}\overrightarrow r }}{{d{t^2}}}\)
11. For an isolated system, the external force acting on the body is zero \(\overrightarrow F = M\frac{d}{{dt}}\left( {{{\overrightarrow v }_{{\text{cm}}}}} \right) = 0\). Where \({\overrightarrow v _{{\text{cm}}}} = \) constant. i.e., the Centre of mass of an isolated system moves with uniform velocity along a straight-line path.

Rotational Motion – Some Important Terminologies

1) Angular Displacement: The angle described by the position vector \(\overrightarrow r\) about the axis of rotation is called angular displacement.

Angular displacement \((\theta ) = \frac{{{\text{Linear}}\,{\text{displacement (}}s)}}{{{\text{Radius (}}r)}}\)
Unit: radian
Dimension: \([{M^0}{L^0}{T^0}]\)
Vector form: \(\overrightarrow S = \overrightarrow \theta \times \overrightarrow r \)

Note: The angular displacement of a body is a vector quantity whose direction is given by the right-hand clasp rule. It is also an axial vector. For anti-clockwise rotation, the direction \(\theta\) is perpendicular to the plane, outward and along the axis of rotation and vice-versa.

2) Angular Velocity: Angular velocity \(\omega\) is the angular displacement per unit of time. Let a particle move from \(P\) to \(Q\) in time \(\Delta t\), \(\omega = \frac{{\Delta \theta }}{{\Delta t}}\) where \(\Delta \theta\) is the angular displacement.

Instantaneous angular velocity \(\omega = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta \theta }}{{\Delta t}} = \frac{{d\theta }}{{dt}}\)
Average angular velocity \({\omega _{av}} = \frac{{{\text{total angular displacement}}}}{{{\text{total time}}}} = \frac{{{\theta _2} – {\theta _1}}}{{{t_2} – {t_1}}}\)
Unit: Radian/sec
Dimension: \([{M^0}{L^0}{T^{ – 1}}]\) which is the same as that of frequency
Vector form: \(\overrightarrow v = \overrightarrow \omega \times \overrightarrow r \) [where \(\overrightarrow v =\) linear velocity, \(\overrightarrow r =\) radius vector]

Note: Angular velocity \(\overrightarrow \omega\) is an axial vector whose direction is normal to the rotational plane, and the right-hand screw rule gives its direction.

3) Angular Acceleration: Angular acceleration \(\alpha\) is the time rate of change of angular velocity.
If the particle has an angular velocity \(\omega_1\) at a time \(t_1\) and angular velocity \(\omega_2\) at a time \(t_2\) then,
Angular acceleration \(\overrightarrow \alpha = \frac{{{{\overrightarrow \omega }_2} – {{\overrightarrow \omega }_1}}}{{{t_2} – {t_1}}}\)
Instantaneous angular acceleration \(\overrightarrow \alpha = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta \omega }}{{\Delta t}} = \frac{{d\overrightarrow \omega }}{{dt}} = \frac{{{d^2}\overrightarrow \theta }}{{d{t^2}}}\).
Average angular acceleration \({\alpha _{av}} = \frac{{{\omega _2} – {\omega _1}}}{{{t_2} – {t_1}}}\).
Unit: \({\text{rad/se}}{{\text{c}}^{\text{2}}}\)
Dimension: \([{M^0}{L^0}{T^{ – 2}}]\)

Note:
1. If \(\alpha = 0\) the circular or rotational motion is said to be uniform.
2. Relation between angular acceleration and linear acceleration \(\overrightarrow a = \overrightarrow \alpha \times \overrightarrow r \).

Equations of Linear Motion and Rotational Motion

Moment of Inertia

Moment of inertia in rotational motion plays the same role as mass plays in rectilinear motion. A body tends to oppose any change in its state of rest or of uniform rotation.

1) The moment of inertia of a particle is given by \(I = mr^2\), where \(r\) is the perpendicular distance of the particle from the rotational axis.
Dimension: \([M{L^2}{T^0}]\)
S.I. unit: \({\text{kg}}\,{{\text{m}}^2}\)
2) Moment of inertia \(I\) of a system made up of several particles distributed discretely
\(I = {m_1}r_1^2 + {m_2}r_2^2 + {m_3}r_3^2 + …….\)
3) Moment of inertia of a body of a continuous distribution of mass, treating the element of mass \(dm\) at the position \(r\) as particle
\(dI = dm\,{r^2}\) i.e., \(I = \int {{r^2}dm} \)

4) The moment of inertia depends on mass, the position of the axis of rotation and mass distribution.
5) Moment of inertia doesn’t depend on angular velocity, torque, angular acceleration, angular momentum and rotational kinetic energy.
6) It is not a vector as direction cannot be specified exactly and not a scalar as it has a different magnitude in different directions and orientations. It is a tensor quantity.
7) In the case of a solid and hollow body of the same mass, shape and radius for a given axis, the moment of inertia of a hollow body is greater than that for the solid body because it depends upon the mass distribution.

Radius of Gyration

The radius of gyration \((k)\) of a body(system of particles) about a particular axis is the perpendicular or normal distance of a point from the axis, where we can imagine the whole mass of the body is concentrated, and the body will have the same moment of inertia as it has with the actual mass distribution.

When the square of the radius of gyration of the body is multiplied by the body’s mass, it gives the moment of inertia about the given axis.
\(I = M{k^2}\) or \({\text{ }}k = \sqrt {\frac{I}{M}} \).
Here \(k\) is called the radius of gyration.
From the formula of discrete distribution
\(I = m{r_1}^2 + m{r_2}^2 + m{r_3}^2 + ……. + mr_n^2\)
If \({m_1} = {m_2} = {m_3} = ………….. = m\) then
\(I = m({r_1}^2 + {r_2}^2 + {r_3}^2 + ……….r_n^2)\) ………….(i)
From the definition of the radius of gyration,
\(I = M{k^2}\) ………………..(ii)
By equating (i) and (ii)
\(M{k^2} = m({r_1}^2 + {r_2}^2 + {r_3}^2 + ………… + r_n^2)\)
\(nm{k^2} = m({r_1}^2 + {r_2}^2 + {r_3}^2 + ………. + r_n^2)\) [As \(M = nm\)]
\(\therefore \,k = \sqrt {\frac{{{r_1}^2 + {r_2}^2 + {r_3}^2 + ……….. + r_n^2}}{n}} \)
Therefore, the radius of gyration of a body about a given axis is equal to the root mean square distance of the constituent particles of the body from the given axis.

Theorem of Parallel Axes

Moment of inertia of a body (system of particles) about a particular axis of rotation \(I\) is equal to the sum of the moment of inertia of the body about an axis parallel to a given axis and passing through the Centre of mass of the body \(I_G\) and \(Ma^2\) where \(M\) is the mass of the body and \(a\) is the perpendicular distance between the two axes.

\(I = {I_G} + M{a^2}\)

Example: Moment of inertia of a circular disc about an axis through its Centre of mass and perpendicular to the plane is \(\frac{1}{2}M{R^2}\), so moment of inertia about an axis through its tangent and perpendicular to the plane will be

\(I = {I_G} + M{a^2}\)
\(I = \frac{1}{2}M{R^2} + M{R^2}\)
\(\therefore \,I = \frac{3}{2}M{R^2}\).

Theorem of Perpendicular Axes

According to this theorem, the sum of the moment of inertia of a plane lamina about two mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of first two axes.

\({I_z} = {I_x} + {I_y}\)
Example: Moment of inertia of a circular disc about an axis through its Centre of mass and perpendicular to its plane is \(\frac{1}{2}M{R^2}\), so if the disc is in \(x – y\) plane then by the theorem of perpendicular axes
i.e., \({I_z} = {I_x} + {I_y}\)

\( \Rightarrow \frac{1}{2}M{R^2} = 2{I_D}\) [As the ring is a symmetrical body so \({I_x} = {I_y} = {I_D}\)]
\( \Rightarrow {I_D} = \frac{1}{4}M{R^2}\).

Torque

If a pivoted, hinged or suspended body tends to rotate under the action of a force, it is said to be acted upon by a torque. The turning effect of a force about the axis of rotation is called the moment of force or torque due to the force.

Consider a particle rotating in a \(xy\) plane about the origin \((0,\,0)\) under the effect of an unbalanced force \(\overrightarrow F\).

At any instant, if the position vector of the particle is \(\overrightarrow r\) then,
Torque \(\overrightarrow \tau = \overrightarrow r \times \overrightarrow F \)
\(\tau = r\,F\sin \phi \)
(Where \(\phi \) is the angle between \(\overrightarrow r\) and \(\overrightarrow F\))
Unit: Newton-metre (MKS system) and Dyne-cm (CGS system)
Dimension: \([M{L^2}{T^{ – 2}}]\)
1) Torque is an axial vector. i.e., its direction is perpendicular to the plane containing vector \(\overrightarrow r\) and \(\overrightarrow F\), according to the right-hand screw rule or clasp rule. If the sense of rotation is anti-clockwise, then the torque direction is perpendicular to the plane, outward through the axis of rotation.
2) Rectangular components of force
\({\overrightarrow F _r} = F\cos \phi = \) radial component of force, \({\overrightarrow F _\phi } = F\sin \phi = \) transverse component of force
As \(\tau = r\,F\,\sin \phi \)
Or \(\tau = r\,{F_\phi } = \) (position vector) \(\times\) (transverse component of force)
The torque is given by the product of the transverse (normal) component of a force and its perpendicular distance from the fixed axis of rotation, i.e., torque is due to the transverse component of force only.
3) As \(\tau = r\,F\,\sin \phi \)
Or \(\tau = F(r\sin \phi ) = Fd\) [As \(d = r\sin \phi \) from the figure]
i.e. Torque \(=\) Force \(\times\) Perpendicular distance
Torque is also called a ‘moment of force’ and \(d\) is called a moment or lever arm.
4) Maximum and minimum torque: As \(\overrightarrow \tau = \overrightarrow r \times \overrightarrow F \) or \(\tau = r\,F\sin \phi \)

5) For a given angle and force, the magnitude of torque depends on \(r\). The more the value of \(r\); the more the torque will be and the easier it is to rotate the body.
Example:
(i) Handles are provided near the edge of the Planck of the door.
(ii) The handle of the screwdriver is taken thick.
(iii) In villages handle of the flour mill is placed near the circumference.
(iv) The handle of the hand pump is kept long.
(v) The arm of the wrench used for opening the tap is kept long.
6) If numerous forces act upon a body, the total torque is the vector sum of each torque.
\(\overrightarrow \tau = {\overrightarrow \tau _1} + {\overrightarrow \tau _2} + {\overrightarrow \tau _3} + ……….\)
7) A body is in rotational equilibrium if the resultant torque acting on it is zero, i.e., \(\sum {\overrightarrow \tau } = 0\).
8) In the case of beam balance or see-saw, the system will be in rotational equilibrium if,

\({\overrightarrow \tau _1} + {\overrightarrow \tau _2} = 0\) Or \({F_1}{l_1} – {F_2}{l_2} = 0\)
\(\therefore \,{F_1}{l_1} = {F_2}{l_2}\)
However if, \({\overrightarrow \tau _1} > {\overrightarrow \tau _2}\) LHS will move downwards and if \({\overrightarrow \tau _1} < {\overrightarrow \tau _2}\) RHS will move downward. The system will not be in rotational equilibrium.

Analogy Between Translatory Motion and Rotational Motion

Angular Momentum

The turning momentum of a particle about the axis of rotation is called the particle’s angular momentum.
Consider \(\overrightarrow P\) is the linear momentum of a particle and \(\overrightarrow r\) its position vector from the point of rotation.

Angular momentum, \(\overrightarrow L = \overrightarrow r \times \overrightarrow P \)
\(\overrightarrow L = r\,P\sin \phi \,\widehat n\)
Angular momentum \(\overrightarrow L\) is always an axial vector, i.e. direction along the axis of rotation and directed perpendicular to the plane of rotation.
SI. Unit: \({\text{kg}}\,{{\text{m}}^{\text{2}}}\,{{\text{s}}^{{\text{-1}}}}\) or \({\text{J sec}}{\text{.}}\)
Dimension: \({\text{[}}M{L^2}{T^{{\text{ – 1}}}}{\text{]}}\) and it is similar to Planck’s constant \((h)\).
1) In Cartesian co-ordinates if \(\overrightarrow r = x\widehat i + y\widehat j + z\widehat k\) and \(\overrightarrow P = {P_x}\widehat i + {P_y}\widehat j + {P_z}\widehat k\)
Then \(\overrightarrow L  = \overrightarrow r  \times \overrightarrow P  = \left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
  x&y&z \\
  {{P_x}}&{{P_y}}&{{P_z}}
\end{array}} \right|\)
\( = {\text{(y}}{P_z} – z{P_y}{\text{)}}\widehat i – (x{P_z} – z{P_x})\widehat j + (x{P_y} – y{P_x})\widehat k\)
2) As it is clear from the figure radial component of momentum \({\overrightarrow P _r} = P\cos \phi \)

Transverse component of momentum \({\overrightarrow P _\phi } = P\sin \phi \)
So the magnitude of angular momentum \(L = r\,P\sin \phi \)
\(L = r\,{P_\phi }\)
Angular momentum \(=\) Position vector \(\times\) Transverse component of angular momentum, i.e., the radial component of linear momentum doesn’t play any role in angular momentum.
3) The magnitude of angular momentum \(L = P\,{\text{(}}r\sin \phi ) = L = Pd\) [As \(d = r\sin \phi \) from the figure.]
4 Angular momentum \(=\) (Linear momentum) \( \cdot \) (Perpendicular distance of the line of action of a force from the axis of rotation)
4) Maximum and minimum angular momentum: We know \(\overrightarrow L = \overrightarrow r \times \overrightarrow P \)
\(\therefore \,\overrightarrow L = m\,[\overrightarrow r \times \overrightarrow v ]\, = m\,v\,r\sin \phi = \,P\,r\sin \phi \) [As \(\overrightarrow P = m\overrightarrow v \)]

Law of Conservation of Angular Momentum

Newton’s Second Law of rotational motion can be given by \(\overrightarrow \tau = \frac{{d\overrightarrow L }}{{dt}}\)
If the net torque on a body is zero, then \(\frac{{d\overrightarrow L }}{{dt}} = 0\)
i.e., \(\overrightarrow L = {\overrightarrow L _1} + {\overrightarrow L _2} + {\overrightarrow L _3} + …….. = \) constant.
Angular momentum of a system (maybe body or particle) remains constant (same) if the resultant torque is zero.
As \(L = I\omega \) so if \(\overrightarrow \tau = 0\), then \( I \omega =\) constant \(\therefore \,I \propto \frac{1}{\omega }\)
Since angular momentum \( I \omega\) remains constant when \(I\) decreases, angular velocity \(\omega\) increases and vice-versa.

Examples for Law of Conservation of Angular Momentum

1) The angular frequency of revolution \(\omega\) of a planet \(P\) around the sun in an elliptical orbit increases when the planet comes nearer to the sun and vice-versa because when the planet comes nearer to the sun, its moment of inertia \(I\) decreases, therefore, \(\omega\) increases.
2) A circus acrobat or stunt master performs feats (stunts) involving spin by bringing his arms and legs closer to his body or vice-versa. When he brings his arms and legs closer to the body, his moment of inertia \(I\) decreases and \(\omega\) increases.
3) A person carrying a heavy load in his hands and standing on a rotating platform can change the platform’s speed using the Angular momentum conservation principle. When the person suddenly folds his arms, his moment of inertia decreases and the angular speed \(\omega\) increases.

4) A diver performs different somersaults by jumping from a high diving board, keeping his legs and arms stretched out first and then curling his body.
5) Change in the radius of the earth affects its time-period
Angular momentum of the earth \(L = I \omega =\) constant
\(L = \frac{2}{5}M{R^2} \times \frac{{2\pi }}{T} =\) constant
\(\therefore \,T \propto {R^2}\) [if M remains constant]
If R becomes half, then the Time period will become one-fourth, i.e \(\frac{{24}}{4} = 6\,{\text{hrs}}\).

Pure Rotational Motion – Solved Examples

Q.1. The coordinates of the positions of particles of mass \(7,\,4\) and \(10\,{\text{gm}}\) are \({\text{(1,}}\,{\text{5,}}\, – {\text{3),}}\,{\text{(2,}}\,5,\,7{\text{)}}\) and \({\text{(3,}}\,{\text{3,}}\, – {\text{1)}}\,{\text{cm}}\), respectively. The position of the Centre of mass(COM) of the system would be
(a) \(\left( { – \frac{{15}}{7},\,\frac{{85}}{{17}},\,\frac{1}{7}} \right)\,{\text{cm}}\)
(b) \(\left( {\frac{{15}}{7},\, – \frac{{85}}{{17}},\,\frac{1}{7}} \right)\,{\text{cm}}\)
(c) \(\left( {\frac{{15}}{7},\,\frac{{85}}{{21}},\, – \frac{1}{7}} \right)\,{\text{cm}}\)
(d) \(\left( {\frac{{15}}{7},\,\frac{{85}}{{21}},\,\frac{7}{3}} \right)\,{\text{cm}}\)
Ans: (c) \({m_1} = 7\,{\text{gm}}\), \({m_2} = 4\,{\text{gm}}\), \({m_3} = 10\,{\text{gm}}\) and
\(\overrightarrow {{r_1}} = (\hat i + 5\hat j – 3\hat k),\,{r_2} = (2i + 5j + 7k)\), \({r_3} = (3\hat i + 3\hat j – \hat k)\)
The position vector of the Centre of mass
\(\overrightarrow r = \frac{{7(\hat i + 5\hat j – 3\hat k) + 4(2\hat i + 5\hat j + 7\hat k) + 10(3\hat i + 3\hat j – \hat k)}}{{7 + 4 + 10}}\) \( = \frac{{(45\hat i + 85\hat j – 3\hat k)}}{{21}}\)
\( \Rightarrow \overrightarrow r = \frac{{15}}{7}\hat i + \frac{{85}}{{21}}\hat j – \frac{1}{7}\hat k\).
So, coordinates of Centre of mass \(\left[ {\frac{{15}}{7},\,\frac{{85}}{{21}},\,\frac{{ – 1}}{7}} \right]\).

Q.2. The velocities of three particles of masses \(20\,{\rm{g}},\,30{\rm{g}}\) and \(50{\rm{g}}\) are \(10\overrightarrow i,\,10\overrightarrow j\) and \(10 \overrightarrow k\), respectively. The velocity of the Centre of mass(COM) of the three particles is
(a) \(2\overrightarrow i + 3\overrightarrow j + 5\overrightarrow k \)
(b) \(10(\overrightarrow i + \overrightarrow j + \overrightarrow k) \)
(c) \(20\overrightarrow i + 30\overrightarrow j + 5\overrightarrow k \)
(d) \(2\overrightarrow i + 30\overrightarrow j + 50\overrightarrow k \)
Ans: (a) Velocity of Centre of mass
\({v_{cm}} = \frac{{{m_1}{v_1} + {m_2}{v_2} + {m_3}{v_3}}}{{{m_1} + {m_2} + {m_3}}}\) \( = \frac{{20 \times 10\hat i + 30 \times 10\hat j + 50 \times 10\hat k}}{{100}}\) \( = 2\hat i + 3\hat j + 5\hat k\).

Q.3. The angular velocity of seconds hand of a watch will be
(a) \(\frac{\pi }{{60}}\,{\text{rad/sec}}\)
(b) \(\frac{\pi }{{30}}\,{\text{rad/sec}}\)
(c) \(60\,\pi \,{\text{rad/sec}}\)
(d) \(30\,\pi \,{\text{rad/sec}}\)
Ans: (b) We know that the second’s hand completes its revolution \((2\pi)\) in \(60\) sec
\(\therefore \,\omega = \frac{\theta }{t} = \frac{{2\pi }}{{60}} = \frac{\pi }{{30}}{\text{rad}}{\text{/}}{\text{sec}}\).

Q.4. The moment of inertia of a uniform circular disc about diameter is \(I\). What will be the moment of inertia about an axis perpendicular (normal) to its plane and passing through a point on its rim?
(a) \(5 I\)
(b) \(6 I\)
(c) \(3 I\)
(d) \(4 I\)
Ans: (b) Moment of inertia of disc about a diameter \( = \frac{1}{4}M{R^2} = I\) (given)
\(\therefore \,M{R^2} = 4I\)
Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim
\( = \frac{3}{2}M{R^2} = \frac{3}{2}(4I) = 6I\).

Q.5. The moment of inertia of a solid sphere of radius \(R\) and density \(\rho\) about its diameter is
(a) \(\frac{{105}}{{176}}{R^5}\rho \)
(b) \(\frac{{105}}{{176}}{R^2}\rho \)
(c) \(\frac{{176}}{{105}}{R^5}\rho \)
(d) \(\frac{{176}}{{105}}{R^2}\rho \)
Ans: (c) Moment of inertia of a sphere about its diameter \(I = \frac{2}{5}M{R^2}\) \( = \frac{2}{5}\left( {\frac{4}{3}\pi {R^3}\rho } \right){R^2}\)
[As \(M = V\rho = \frac{4}{3}\pi {R^3}\rho \)]
\(I = \frac{{8\pi }}{{15}}{R^5}\rho = \frac{{8 \times 22}}{{15 \times 7}}{R^5}\rho = \frac{{176}}{{105}}{R^5}\rho \).

Q.6. A force of \((2\hat i – 4\hat j + 2\hat k)\,{\text{N}}\) acts at a point \((3\hat i + 2\hat j – 4\hat k)\) metre from the origin. The magnitude of torque is
(a) Zero
(b) \(24.4\;{\rm{N}}\,{\rm{m}}\)
(c) \(0.244\;{\rm{N}}\,{\rm{m}}\)
(d) \(2.444\;{\rm{N}}\,{\rm{m}}\)
Ans: \(\overrightarrow F = (2\hat i – 4\hat j + 2\hat k)\,{\text{N}}\) and \(\overrightarrow r = (3\hat i + 2\hat j – 4\hat k)\) meter
Torque \(\overrightarrow \tau \, = \overrightarrow r \times \overrightarrow F \)
\( = \left| {\begin{array}{*{20}{r}}
{\hat i}&{\hat j}&{\hat k} \
3&2&{ – 4} \
2&{ – 4}&2
\end{array}} \right|\)
\( \Rightarrow \overrightarrow \tau = – 12\hat i – 14\hat j – 16\hat k\)
And \(|\overrightarrow \tau | = \sqrt {{{( – 12)}^2} + {{( – 14)}^2} + {{( – 16)}^2}} = 24.4\,{\text{N}}\,{\text{m}}\).

Q.7. A thin circular ring of mass \(M\) and radius \(R\) rotates about its axis with a constant angular speed. Four objects each of mass \(m\) are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular speed of the ring will be
(a) \(\frac{{M\omega }}{{M + 4m}}\)
(b) \(\frac{{(M + 4m)\omega }}{M}\)
(c) \(\frac{{(M – 4m)\omega }}{{M + 4m}}\)
(d) \(\frac{{M\omega }}{{4m}}\)
Ans: (a) Initial angular momentum of ring \( = I\omega = M{R^2}\omega \)
If four object each of mass \(m\) and kept gently to the opposite ends of two perpendicular diameters of the ring, then final angular momentum \( = (M{R^2} + 4m{R^2})\omega ‘\)
By the conservation of angular momentum
Initial angular momentum \(=\) Final angular momentum
\(M{R^2}\omega = (M{R^2} + 4m{R^2})\omega ‘ \Rightarrow \omega ‘ = \left( {\frac{M}{{M + 4m}}} \right)\omega \).

Summary

1. A rigid body is one in which there is no change in shape and size on the application of deforming forces. The linear distance between any two points in the rigid body always remains the same.
2. Pure Rotational motion is the motion of a body about a fixed axis such that all the particles constituting it undergoes circular motion about that fixed axis.
3. The Centre of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there.
4. Moment of inertia is the tendency of a body due to which it opposes any change in its state of rest or of uniform rotational motion.
5. The angular momentum of a system of particles (maybe particle or body) remains constant if the resultant torque acting on it is zero. This law is called as law of conservation of angular momentum.

Pure Rotational motion – FAQs

Q.1. Why is the concept of the Centre of mass (COM) required?
Ans: Centre of mass (COM) describes the behaviour of a macroscopic body in terms of the laws developed for the tiny bodies.

Q.2. Does the Centre of mass (COM) of a body necessarily lie inside the body?
Ans: No. For example, the COM of a ring lies in its hollow portion.

Q.3. Name the physical quantity that is also called the moment of force. What are the factors on which it depends?
Ans: The moment of force is also called torque. It depends on (i) the magnitude of force, (ii) the perpendicular (normal) distance of the line of action of a force from the fixed axis of rotation.

Q.4. A body is in pure rotational motion. Is it always necessary that a torque be acting on it?
Ans: No, torque is required only for angular acceleration.

Q.5. Why is a force applied at right angles to the heavy door at the outer edge while closing or opening it?
Ans: Torque \(\tau = rF\sin \theta \). For a force applied perpendicular to the outer edge of the door, both \(\sin \theta = \sin {90^0} = 1 = \) maximum and \(r\) are maximum. Hence the torque produced is maximum.

Q.6. If net external torque acting on a body is zero, which physical quantity remains conserved?
Ans: Angular momentum

Q.7. What is the physical significance of the concept of moment of inertia?
Ans: The moment of inertia plays the role of mass in rotatory motion. It gives a measure of inertia (resistance to change) in rotational motion.

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