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It is defined as the distance from the axis of rotation to a point where the total mass of the body is supposed to be concentrated so that the moment of inertia about the axis may remain the same. It is denoted by K. In terms of radius of gyration, the moment of inertia of the body of mass M is given as
$$ I = MK^2 \dots (i) $$
Suppose a body consists of n particles each mass m. Let r_{1}, r_{2}, r_{3}, …. r_{n} be their perpendicular distances from the axis of rotation. Then, the moment of inertia I of the body about the axis of rotation is
\begin{align*} I &= m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + \dots + m_nr_n^2 \\ \text {If all the particles are of same mass m, then} \\ I &= m(r_1^2 + r_2^2 + r_3^2 + \dots + r_n^2) \\ &=\frac { mn(r_1^2 + r_2^2 + r_3^2 + \dots + r_n^2)}{n} \\ \text {Since mn} = \text {M, total mass of the body,} \\ \therefore I &= M\left (\frac { r_1^2 + r_2^2 + r_3^2 + \dots + r_n^2}{n}\right ) \dots (ii) \\ \text {from equation} (i) \text {and equation} (ii), \text {we have} \\ MK^2 &= M\left (\frac { r_1^2 + r_2^2 + r_3^2 + \dots + r_n^2}{n}\right ) \\ K &= \sqrt {M \frac { r_1^2 + r_2^2 + r_3^2 + \dots + r_n^2}{n}} \\ &= \text {root mean square distance of the particles from the axis of rotation} \end{align*}
Therefore, the radius of gyration of a body about a given axis may also defined as the root mean square distance of the various particles of the body from the axis of rotation.
The moment of inertia of uniform thin rod of mass M and length l about an axis through its centre and perpendicular to its length is given by
\begin{align*} I &= \frac {Ml^2}{12} \dots (iii) \\ \text {If K is the radius of gyration of the rod about the axis, then we have} \\ I &= MK^2 \dots (iv) \\ \text {from equation} (i) \text {and equation} (ii), \text {we have} \\ MK^2 &= \frac {Ml^2}{\sqrt {12}} \\ \therefore K &= \frac {l}{12} \end{align*}
The moment of inertia for a solid sphere of radius R and mass M is given by
\begin{align*} I &= \frac 25 MR^2 \\ \text {If K is the radius of gyration of the solid sphere, then} \\ I &= MK^2 \\ \text {or,} MK^2 &= \frac 25 MR^2 \\ \text {or,} K &= \sqrt {\frac 25} R \\ \end{align*}
The turning effect of a force in a body is called torque or moment of force. Its magnitude can be calculated by taking the product of the force and its perpendicular distance from the axis of rotation. Generally, it is denoted by ().
\begin{align*} \text {Torque} &= \text {force} \times \text {perpendicular distance of the force from the axis of rotation} \\ \text {or,} \: \tau &= rF \\ \end{align*}
Torque is the vector quantity. So in terms of a vector
$$ \vec \tau = \vec r \times \vec F $$
whose direction id along the cross product of two vectors as shown in the figure.
If the body rotates in clockwise direction, then the torque applied on the body is said to be clockwise torque. On the other hand, the torque applied on a body is said to be anticlockwise torque, if it rotates the body in an anticlockwise direction.
In CGS-system, its unit is dyne cm and in SI-units, its unit is Nm. The dimensional formula of torque is [ML^{2}T^{-2}].
Let us suppose that a body rotating about an axis YY’ under the action of a constant torque(). Suppose the body consists of n particles of masses m_{1}, m_{2}, m_{3}, …., m_{n} at distance r_{1}, r_{2}, r_{3}, ….. r_{n} from the axis of rotation respectively. The torque will produce a constant angular acceleration (α) in each particle. Since the particle of mass m_{1} follows a circular path of radius r_{1}, the magnitude of the linear acceleration of this particle is
\begin{align*} a_1 = r_1\alpha \\ \text {The net external force acting on this particle is} \\ F_1 &= m_1a_1 = m_1(r_1\alpha ) = m_1r_1\alpha \\ \text {The magnitude of torque acting on this particle due to the force} \\ \tau _1 &= F_1r_1 \\ &= (m_1r_1\alpha ) r_1 \\ &= m_1r_1^2 \alpha \\ \end{align*}
Similarly, the magnitude of torque on the particles of masses m_{2}, m_{3}, …. are m_{2}r_{2}^{2}α , m_{3}r_{3}^{2}α …. respectively. Since the torque acting on the body is equal to the sum of the individual torques acting on the constituent particles, so
\begin{align*} \tau &= m_1r_1^2 \alpha + m_2r_2^2 \alpha + m_3r_3^2 \alpha + \dots \\ &= (m_1r_1^2 + m_2r_2^2 + m_3r_3^2 + \dots ) \alpha \\ &= (\sum m_ir_i^2) \alpha \\ \therefore \tau &= I\alpha \end{align*}
where, \(I = \sum m_ir_i^2 \) is the moment of inertia of the body about the axis YY’. This is the relation between moment of inertia of a body and the torque.
\begin{align*} if \: \alpha = 1, \text {then} \: \tau = I \times 1 \\ \therefore I &= \tau \end{align*}
Hence, the moment of inertia of a body about a given axis is equal to the torque required to produce unit angular acceleration in the body about that axis.
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ASK ANY QUESTION ON Radius of Gyration and Torque
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