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Moment of Inertia

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a week ago

Moment of Inertia Class 11th

Two solid spheres each of mass 2 kg and radius 75 cm are arranged as shown. Find MOI of the system about the given axis.

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alok kumar singh

Contributor-Level 10

$I = (\frac{2}{5} M R^{2} + M R^{2}) * 2$

$= \frac{14}{5} * 2 * \frac{9}{16}$

$= \frac{63}{20}$

= 3.15 kg m²

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a week ago

Moment of Inertia Class 11th

Two particles of masses 5 kg and 10 kg are placed along x-axis at positions x = 1 m and x = 3 m. The moment of inertia of the system about an axis respectively passing through centre of mass and parallel to y-axis is

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alok kumar singh

Contributor-Level 10

Distance of CM from 5 kg = 10*2/15 = 4/3m

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2 weeks ago

Moment of Inertia Class 11th

Two circular rings (of same material) are of same thickness. The diameter of first ring is twice that of second. The moment of inertia of first ring as compared to that of second is

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alok kumar singh

Contributor-Level 10

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New answer posted

2 weeks ago

Moment of Inertia Class 11th

Moment of inertia of a system made from three uniform rods each of mass m and length l joined to each other, as shown in the figure about one end O and perpendicular to its plane will be

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Raj Pandey

Contributor-Level 9

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New answer posted

3 weeks ago

Moment of Inertia Class 11th

The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I1. The same rod is into a ring and its moment of inertia about a diameter is I2. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ $\frac{_{}}{_{}} \frac{^{}}{}$ then the value of x will be__________.

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} * \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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New answer posted

3 weeks ago

Moment of Inertia Class 11th

The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.

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alok kumar singh

Contributor-Level 10

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} * \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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New answer posted

3 weeks ago

Moment of Inertia Class 11th

A square of mass m is made of thin wire with sides of length a. Calculate the rotational inertia about the z-axis.

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Vishal Baghel

Contributor-Level 10

$I = \frac{m}{4} * \frac{a^{2}}{3} * 2 + \frac{m}{4} * (\frac{a^{2}}{12} + {(\frac{a}{2})}^{2} + a^{2}) * 2$

$= \frac{m a^{2}}{6} + \frac{2}{3} m a^{2} = \frac{m a^{2} + 4 m a^{2}}{6} = \frac{5 m a^{2}}{6}$

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New answer posted

3 weeks ago

Moment of Inertia Class 11th

Three points masses $m, 2 m$ and 3m are located at the vertices of an equilateral triangle of side ' a '. The moment of inertia of the system about an axis along the altitude of the triangle passing through m is $\frac{N}{20} m a^{2}$ where N is an integer. The value of N is

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Raj Pandey

Contributor-Level 9

$I_{A B} = m (0)^{2} + 2 m {(\frac{a}{2})}^{2} + 3 m {(\frac{a}{2})}^{2}$

$\begin{array}{l} I_{A B} = \frac{m a^{2}}{2} + \frac{3 m a^{2}}{4} = \frac{5 m a^{2}}{4} \\ I_{A B} = \frac{25 m a^{2}}{20} = \frac{N}{20} m a^{2} & ∴ N = 25 \end{array}$

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New answer posted

4 weeks ago

Moment of Inertia Class 11th

Moment of inertia (M.I) of four bodies, having same mass and radius, and reported as:

I₁ = M.I. of thin circular ring about its diameter

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

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alok kumar singh

Contributor-Level 10

$l_{1} = \frac{1}{2} m r^{2}$

$\begin{array}{l} l_{2} = \frac{1}{2} m r^{2} \\ l_{3} = \frac{1}{2} m r^{2} \\ l_{4} = \frac{2}{5} m r^{2} \end{array}$

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a month ago

Moment of Inertia Class 11th

Consider a uniform wire of mass M and length L. It is bent into a semi-circle. Its moment of inertia about a line perpendicular to the plane of the wire passing through centre is :

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