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Chapter 5: Problem 607

The M.I of a disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about an axis passing through the centre \(\mathrm{O}\) and perpendicular to the plane of disc is \(\left(\mathrm{MR}^{2} / 2\right)\). If one quarter of the disc is removed the new moment of inertia of disc will be..... \(\\{\mathrm{A}\\}\left(\mathrm{MR}^{2} / 3\right)\) \(\\{B\\}\left(M R^{2} / 4\right)\) \(\\{\mathrm{C}\\}(3 / 8) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(3 / 2) \mathrm{MR}^{2}\)

Short Answer

Expert verified
The new moment of inertia of the disc after one quarter has been removed is \(I' = \frac{3MR^2}{8}\).

Step by step solution

01

Moment of Inertia of Full Disc

The initial moment of inertia (I) of the full disc with mass M and radius R is given by: \(I = \frac{MR^2}{2}\)
02

Determine the Mass and Radius of the Quarter Disc

Since one quarter of the disc is removed, the mass (m) and radius (r) of the removed quarter disc are: \(m = \frac{M}{4}\) (1/4th of the mass) \(r = R\) (radius remains the same)
03

Moment of Inertia of Quarter Disc

The moment of inertia (i) of the removed quarter disc can be calculated using the same formula: \(i = \frac{mr^2}{2}\) Substitute the values of m and r: \(i = \frac{\left(\frac{M}{4}\right)R^2}{2}\)
04

Simplify the Moment of Inertia of Quarter Disc

Simplify the expression to get the moment of inertia of the quarter disc: \(i = \frac{MR^2}{8}\)
05

New Moment of Inertia of the Disc

The new moment of inertia (I') of the remaining disc can be obtained by subtracting the quarter disc's moment of inertia from the full disc's moment of inertia: \(I' = I - i\) substitute the values: \(I' = \frac{MR^2}{2} - \frac{MR^2}{8}\)
06

Simplify the New Moment of Inertia of the Disc

Simplify the expression to get the new moment of inertia of the disc: \(I' = \frac{4MR^2 - MR^2}{8}\) \(I' = \frac{3MR^2}{8}\)
07

Match with the Correct Option

Now, match the obtained new moment of inertia with the given options: The new moment of inertia of the disc is \(\frac{3MR^2}{8}\), which matches with option C. Therefore, the correct answer is option C. \(I' = \frac{3MR^2}{8}\).

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Most popular questions from this chapter

If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(\mathrm{T}\) is..... \(\\{\mathrm{A}\\}\left(\pi \mathrm{MR}^{3} / \mathrm{T}\right)\) \(\\{\mathrm{B}\\}\left(\operatorname{MR}^{2} \pi / \mathrm{T}\right)\) \(\\{C\\}\left(2 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\) \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\)

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 \(\\{\mathrm{A}\\} 1: 1\) \(\\{\mathrm{B}\\} 2: 1\) \(\\{C\\} 1: 2\) \(\\{\mathrm{D}\\} 4: 1\)

Statement \(-1\) - Friction is necessary for a body to roll on surface. Statement \(-2\) - Friction provides the necessary tangential force and torque. \(\\{\mathrm{A}\\}\) 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 \(-1\). \(\\{\mathrm{C}\\}\) Statement \(-1\) is true, statement \(-2\) is false \\{D \\} Statement- 2 is false, statement \(-2\) is true

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None
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