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Chapter 5: Problem 648

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

Short Answer

Expert verified
The short answer to the question is: The moment of inertia of the disc about an axis passing through its center and normal to the disc is \(I₂ = (1/2)MR²\). The correct option is \(\mathrm{B}\\(1 /2) \mathrm{MR}^{2}\).

Step by step solution

01

Write the given information.

We know that the moment of inertia, I₁, of the disc about any of its diameter is (1/4)MR², and we need to find the moment of inertia, I₂, about an axis passing through the center of the disc and normal to the disc.
02

Write Parallel Axis Theorem formula

According to the Parallel Axis Theorem: I₂ = I₁ + Md²
03

Substitute given values into the formula.

We know I₁ = (1/4)MR² and d = R/2. Let's substitute these values into the formula: I₂ = (1/4)MR² + M(R/2)²
04

Simplify the equation.

Now, let's simplify the equation: I₂ = (1/4)MR² + M(R²/4) I₂ = MR²(1/4 + 1/4)
05

Solve for I₂.

Finally, let's solve for I₂: I₂ = MR²(1/2) The moment of inertia of the disc about an axis passing through its center and normal to it is: \(\boxed{I₂ = (1/2)MR²}\) Therefore, the correct answer is \(\mathrm{B}\\(1 /2) \mathrm{MR}^{2}\).

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Most popular questions from this chapter

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

A wheel rotates with a constant acceleration of $2.0 \mathrm{rad} / \mathrm{sec}^{2}$ If the wheel start from rest. The number of revolution it makes in the first ten seconds will be approximately. \(\\{\mathrm{A}\\} 8\) \\{B \\} 16 \(\\{\mathrm{C}\\} 24\) \(\\{\mathrm{D}\\} 32\)

A gramophone record of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is rotating with angular speed \(\mathrm{W}\). If two pieces of wax each of mass \(\mathrm{M}\) are kept on it at a distance of \(R / 2\) from the centre on opposite side then the new angular velocity will be..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)
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