Theorem of Parallel Axes and Theorem of Perpendicular Axes

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Statement: The moment of inertia of a laminar object about an axis (Z) perpendicular to its plane is the sum of its moment of inertia about two mutually perpendicular axes (X and Y) in its plane, all the three axes being concurrent. Mathematically, $I_z = I_x+I_y$ Proof: i. Consider a rigid laminar object able to rotate about three mutually perpendicular axes X, Y and Z. ii. Axes X and Y are in the plane of the object while the Z axis is perpendicular to it, and all are concurrent at O. iii. Consider a mass element dm located at any point P. PM = y and PN = x are the perpendiculars drawn from P respectively on the X and Y axes. iv. The respective perpendicular distances of point P from X, Y and Z axes will then be y, x and $\sqrt{y^2+x^2}$. v. If $I_x$, $I_y$ and $I_z$ are the respective moment of inertia of the body about x, y and z axes, then $$I_x=\int y^2dm,\quad I_y=\int x^2dm,\quad I_z=\int(y^2+x^2)dm$$ $$\therefore I_z=\int y^2dm+\int x^2dm=I_x+I_y$$ This is the mathematical form of the theorem of perpendicular axes.

2. 2.↑ back to the question

   Statement: The moment of inertia ($I_O$) of an object about any axis is the sum of its moment of inertia ($I_C$) about an axis parallel to the given axis, and passing through the centre of mass and the product of the mass of the object and the square of the distance between the two axes. Mathematically, $I_O = I_C+Mh^2$ Proof: i. Consider an object of mass M. Axis MOP is any axis passing through point O. ii. Axis ACB is passing through the centre of mass C of the object, parallel to the axis MOP, and at a distance h from it ($\therefore h = CO$). iii. Consider a mass element 'dm' located at point D. Perpendicular on OC (produced) from point D is DN. iv. Moment of inertia of the object about the axis ACB is $I_c=\int(DC)^2dm$, and about the axis MOP it is $I_o=\int(DO)^2dm$. $$\therefore I_o=\int(DO)^2dm=\int\left[(DN)^2+(NO)^2\right]dm$$ $$=\int\left[(DN)^2+(NC)^2+2\cdot NC\cdot CO+(CO)^2\right]dm$$ $$=\int\left[(DC)^2+2NC\cdot h+h^2\right]dm\qquad....(\text{using Pythagoras theorem in }\triangle DNC)$$ $$=\int(DC)^2dm+2h\int NC\cdot dm+h^2\int dm$$ Now, $\int(DC)^2dm = I_c$ and $\int dm = M$ v. NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass, $\int NC\cdot dm = 0$ $\therefore I_o = I_c+Mh^2$ This is the mathematical form of the theorem of parallel axes.

3. 3.↑ back to the question

   Parallel axes theorem: [From 1.7, Q2 (Statement only)] Statement: The moment of inertia ($I_O$) of an object about any axis is the sum of its moment of inertia ($I_C$) about an axis parallel to the given axis, and passing through the centre of mass and the product of the mass of the object and the square of the distance between the two axes. Mathematically, $I_O = I_C+Mh^2$ Proof: i. Consider an object of mass M. Axis MOP is any axis passing through point O. ii. Axis ACB is passing through the centre of mass C of the object, parallel to the axis MOP, and at a distance h from it ($\therefore h = CO$). iii. Consider a mass element 'dm' located at point D. Perpendicular on OC (produced) from point D is DN. iv. Moment of inertia of the object about the axis ACB is $I_c=\int(DC)^2dm$, and about the axis MOP it is $I_o=\int(DO)^2dm$. $$\therefore I_o=\int(DO)^2dm=\int\left[(DN)^2+(NO)^2\right]dm$$ $$=\int\left[(DN)^2+(NC)^2+2\cdot NC\cdot CO+(CO)^2\right]dm$$ $$=\int\left[(DC)^2+2NC\cdot h+h^2\right]dm\qquad....(\text{using Pythagoras theorem in }\triangle DNC)$$ $$=\int(DC)^2dm+2h\int NC\cdot dm+h^2\int dm$$ Now, $\int(DC)^2dm = I_c$ and $\int dm = M$ v. NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass, $\int NC\cdot dm = 0$ $\therefore I_o = I_c+Mh^2$ This is the mathematical form of the theorem of parallel axes. Perpendicular axes theorem: [From 1.7, Q1 (Statement only)] Statement: The moment of inertia of a laminar object about an axis (Z) perpendicular to its plane is the sum of its moment of inertia about two mutually perpendicular axes (X and Y) in its plane, all the three axes being concurrent. Mathematically, $I_z = I_x+I_y$ Proof: i. Consider a rigid laminar object able to rotate about three mutually perpendicular axes X, Y and Z. ii. Axes X and Y are in the plane of the object while the Z axis is perpendicular to it, and all are concurrent at O. iii. Consider a mass element dm located at any point P. PM = y and PN = x are the perpendiculars drawn from P respectively on the X and Y axes. iv. The respective perpendicular distances of point P from X, Y and Z axes will then be y, x and $\sqrt{y^2+x^2}$. v. If $I_x$, $I_y$ and $I_z$ are the respective moment of inertia of the body about x, y and z axes, then $$I_x=\int y^2dm,\quad I_y=\int x^2dm,\quad I_z=\int(y^2+x^2)dm$$ $$\therefore I_z=\int y^2dm+\int x^2dm=I_x+I_y$$ This is the mathematical form of the theorem of perpendicular axes.

4. 4.↑ back to the question

   Expression of M.I of uniform disc about an axis passing through its centre and perpendicular to its plane: i. Let, M = mass of disc, R = radius of disc, ZZ' = axis passing through the centre of disc and perpendicular to the plane. ii. The M.I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by, $$I_C=\dfrac{1}{2}MR^2$$

5. 5.↑ back to the question

   [From 1.7, Q2] Statement: The moment of inertia ($I_O$) of an object about any axis is the sum of its moment of inertia ($I_C$) about an axis parallel to the given axis, and passing through the centre of mass and the product of the mass of the object and the square of the distance between the two axes. Mathematically, $I_O = I_C+Mh^2$ Proof: i. Consider an object of mass M. Axis MOP is any axis passing through point O. ii. Axis ACB is passing through the centre of mass C of the object, parallel to the axis MOP, and at a distance h from it ($\therefore h = CO$). iii. Consider a mass element 'dm' located at point D. Perpendicular on OC (produced) from point D is DN. iv. Moment of inertia of the object about the axis ACB is $I_c=\int(DC)^2dm$, and about the axis MOP it is $I_o=\int(DO)^2dm$. $$\therefore I_o=\int(DO)^2dm=\int\left[(DN)^2+(NO)^2\right]dm$$ $$=\int\left[(DN)^2+(NC)^2+2\cdot NC\cdot CO+(CO)^2\right]dm$$ $$=\int\left[(DC)^2+2NC\cdot h+h^2\right]dm\qquad....(\text{using Pythagoras theorem in }\triangle DNC)$$ $$=\int(DC)^2dm+2h\int NC\cdot dm+h^2\int dm$$ Now, $\int(DC)^2dm = I_c$ and $\int dm = M$ v. NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass, $\int NC\cdot dm = 0$ $\therefore I_o = I_c+Mh^2$ This is the mathematical form of the theorem of parallel axes.

6. 6.↑ back to the question

   Given: M = 50 g $=50\times10^{-3}$ kg, R = 2 cm $=2\times10^{-2}$ m, L = 12 cm $=12\times10^{-2}$ m To find: Moment of inertia (I) Formula: $I=M\left(\dfrac{R^2}{4}+\dfrac{L^2}{12}\right)$ Calculation: For a solid cylinder, its M.I. about an axis passing through centre perpendicular to its length, $$I=50\times10^{-3}\left(\dfrac{2^2}{4}+\dfrac{12^2}{12}\right)\times(10^{-2})^2$$ $=50\left(\dfrac{4}{4}+\dfrac{144}{12}\right)\times10^{-7}=50(1+12)\times10^{-7}=650\times10^{-7}=6.5\times10^{-5}$ kg m$^2$ Ans: Moment of inertia of given solid cylinder is $6.5\times10^{-5}$ kg m$^2$.

7. 7.↑ back to the question

   Given: R = 0.1 m, $\rho=6\times10^3$ kg/m$^3$ To find: M.I of sphere about a tangent to its surface ($I_t$) Formula: $I_t=\dfrac{7}{5}MR^2$ Calculation: $I_t=\dfrac{7}{5}MR^2$ But $M=\rho V=\dfrac{4}{3}\pi R^3\rho$ $$\therefore I_t=\dfrac{7}{5}\left(\dfrac{4}{3}\pi R^3\rho\right)R^2=\dfrac{28}{15}\pi R^5\rho=\dfrac{28}{15}\times3.14\times(0.1)^5\times6\times10^3$$ $$=\dfrac{28\times3.14\times6}{15}\times10^{-2}=\dfrac{56\times3.14}{5}\times10^{-2}=35.17\times10^{-2}$$ $\therefore I_t=0.3517$ kg m$^2$ Ans: Moment of inertia of solid sphere about a tangent to its surface is 0.3517 kg m$^2$.

8. 8.↑ back to the question

   Given: R = 0.2 m, $\rho=8000$ kg/m$^3$ To find: Moment of inertia (I) Formulae: i. $I_o=I_c+MR^2 \therefore I_o=\dfrac{7}{5}MR^2$ ii. Mass (M) = volume × density Calculation: From formula (ii), $M=V\rho=\left(\dfrac{4}{3}\pi R^3\right)\rho$ From formula (i), $I=\dfrac{7}{5}\left(\dfrac{4}{3}\pi R^3\rho\right)R^2=\dfrac{28}{15}\pi R^5\rho$ $$=\dfrac{28}{15}\times3.142\times(2\times10^{-1})^5\times8000=\dfrac{28\times3.142\times32\times8}{15}\times10^{-2}$$ $=\{\text{antilog}[\log(28)+\log(3.142)+\log(32)+\log(8)-\log(15)]\}\times10^{-2}$ $=\{\text{antilog}[1.4472+0.4972+1.5051+0.9031-1.1761]\}\times10^{-2}$ $=\{\text{antilog}[3.1765]\}\times10^{-2}=1501\times10^{-2}$ $\therefore I=15.01$ kg m$^2$ Ans: M.I. of the uniform solid sphere about a tangent to its surface is 15.01 kg m$^2$.

9. 9.↑ back to the question

   Radius of gyration about given axis parallel to the axis passing through centre of mass is, $K_P$ = 0.5 m Distance between two axes h = 0.4 m From parallel axes theorem, $I_P=I_C+Mh^2$ $\therefore MK_P^2=MK_C^2+Mh^2$ $\therefore K_C^2=K_P^2-h^2$ $\therefore K_C=\sqrt{K_P^2-h^2}=\sqrt{(0.5)^2-(0.4)^2}=\sqrt{0.25-0.16}=\sqrt{0.09}=0.3$ m Ans: The radius of gyration about axis passing through the centre of mass is 0.3 m.

10. 10.↑ back to the question

    Given: L = 3 m To find: Radius of gyration (K) Formulae: i. $I=\dfrac{1}{3}ML^2$ ii. $I=MK^2$ Calculation: From formulae (i) and (ii), $K^2=\dfrac{L^2}{3}$ $\therefore K=\dfrac{L}{\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$ m Ans: Radius of gyration about the transverse axis passing through one end of the rod is $\sqrt3$ m.

11. 11.↑ back to the question

    Given that, $I_C$ = 2 kg-m$^2$ According to perpendicular axis theorem (sic — parallel axis theorem), $I_P=I_C+MR^2$ In case of sphere, $I_C=\dfrac{2}{5}MR^2$ $\therefore I_P=\dfrac{2}{5}MR^2+MR^2=\dfrac{7}{5}MR^2$ $\therefore \dfrac{2}{5}MR^2=2 \therefore MR^2=5$ $\therefore I_P=\dfrac{7}{5}MR^2=\dfrac{7}{5}\times5=7$ kg-m$^2$ Ans: M.I. about the tangent is 7 kg-m$^2$.

12. 12.↑ back to the question

    M.I. of uniform disc about an axis perpendicular to its length and passing through its centre, $$I=\dfrac{1}{2}MR^2=\dfrac{1}{2}\times10\times(0.6)^2=5\times0.36$$ $\therefore I=1.8$ kg m$^2$

13. 13.↑ back to the question

    Given: R = 10 cm = 0.1 m, $\sigma=2$ kg/m$^2$ To find: Moment of inertia (I) Formulae: i. $\sigma=\dfrac{M}{\pi R^2}$ ii. $I=\dfrac{1}{2}MR^2$ Calculation: From formula (i), $M=(\pi R^2)\sigma$ From formula (ii), $I=\dfrac{1}{2}(\pi R^2)\sigma R^2=\dfrac{1}{2}\pi R^4\sigma=\dfrac{1}{2}\times3.142\times(1\times10^{-1})^4\times2$ $\therefore I=3.142\times10^{-4}$ kg m$^2$ Ans: M.I. of the uniform disc about an axis passing through its centre and perpendicular to its plane is $3.142\times10^{-4}$ kg m$^2$.