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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

One circular rig and one circular disc both are having the same mass and radius. The ratio of their moment of inertia about the axes passing through their centers and perpendicular to their planes, will be \(\\{\mathrm{A}\\} 1: 1\) \(\\{\mathrm{B}\\} 2: 1\) \(\\{C\\} 1: 2\) \(\\{\mathrm{D}\\} 4: 1\)

The moment of inertia of a meter scale of mass \(0.6 \mathrm{~kg}\) about an axis perpendicular to the scale and passing through \(30 \mathrm{~cm}\) position on the scale is given by (Breath of scale is negligible). \(\\{\mathrm{A}\\} 0.104 \mathrm{kgm}^{2}\) \\{B \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)

Statement \(-1-\) A thin uniform rod \(A B\) of mass \(M\) and length \(\mathrm{L}\) is hinged at one end \(\mathrm{A}\) to the horizontal floor initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane, The angular velocity of the rod when its ends \(B\) strikes the floor $\sqrt{(3 g / L)}\( Statement \)-2$ - The angular momentum of the rod about the hinge remains constant throughout its fall to the floor. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement - 1 \\{B \\} Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1\). \\{C\\} Statement \(-1\) is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true

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 cylinder of mass \(\mathrm{M}\) has length \(\mathrm{L}\) that is 3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? \(\\{\mathrm{A}\\} 1\) \(\\{\mathrm{B}\\}(1 / \sqrt{3})\) \(\\{\mathrm{C}\\} \sqrt{3}\) \(\\{\mathrm{D}\\}(\sqrt{3} / 2)\)
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