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Chapter 5: Problem 593

Initial angular velocity of a circular disc of mass \(\mathrm{M}\) is \(\omega_{1}\) Then two spheres of mass \(\mathrm{m}\) are attached gently two diametrically opposite points on the edge of the disc what is the final angular velocity of the disc? \(\\{\mathrm{A}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{B}\\}[(\mathrm{M}+4 \mathrm{~m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{C}\\}[\mathrm{M} /(\mathrm{M}+4 \mathrm{~m})] \omega_{1}\) \\{D\\} \([\mathrm{M} /(\mathrm{M}+2 \mathrm{~m})] \omega_{1}\)

Short Answer

Expert verified
The short answer is: \(\\{C\\} [\mathrm{M} /(\mathrm{M}+4 \mathrm{~m})] \omega_{1}\)

Step by step solution

01

Moment of Inertia of the Circular Disc

To find the moment of inertia of the circular disc, use the following formula: \[I_{disc} = \frac{1}{2}MR^2\] Where I is the moment of inertia, M is the mass of the disc, and R is the radius of the disc.
02

Moment of Inertia of the Attached Spheres

Since the two spheres are attached at the edge of the disc, their moment of inertia can be calculated using the following formula: \[I_{sphere} = mR^2\] Where I is the moment of inertia, m is the mass of a sphere, and R is the radius of the disc. Because there are two spheres, the total moment of inertia of the spheres is: \[I_{spheres} = 2mR^2\]
03

Calculate the Initial Angular Momentum

Using the moment of inertia of the circular disc, we can calculate the initial angular momentum, L1: \[L_{1} = I_{disc} \omega_{1} = \frac{1}{2}MR^2 \omega_{1}\]
04

Calculate the Final Angular Momentum

The final angular momentum, L2, will be the sum of the initial angular momentum of the disc and the angular momentum of the attached spheres: \[L_{2} = (I_{disc} + I_{spheres}) \omega_{2} = (\frac{1}{2}MR^2 + 2mR^2) \omega_{2}\]
05

Apply the Law of Conservation of Angular Momentum

Since L1 = L2, we have: \[\frac{1}{2}MR^2 \omega_{1} = (\frac{1}{2}MR^2 + 2mR^2) \omega_{2}\] Now, we can solve for the final angular velocity, ω2: \[\omega_{2} = \frac{\frac{1}{2}MR^2 \omega_{1}}{\frac{1}{2}MR^2 + 2mR^2}\]
06

Simplify the Expression for ω2

We can simplify the expression for ω2: \[\omega_{2} = \frac{MR^2 \omega_{1}}{MR^2 + 4mR^2} = \frac{M \omega_{1}}{M+4m}\] Comparing this result with the multiple-choice options, the correct answer is: \(\\{C\\} [\mathrm{M} /(\mathrm{M}+4 \mathrm{~m})] \omega_{1}\)

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Most popular questions from this chapter

The angular momentum of a wheel changes from \(2 \mathrm{~L}\) to $5 \mathrm{~L}$ in 3 seconds what is the magnitudes of torque acting on it? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)

Statement \(-1\) -If the cylinder rolling with angular speed- w. suddenly breaks up in to two equal halves of the same radius. The angular speed of each piece becomes \(2 \mathrm{w}\). Statement \(-2\) - If no external torque outs, the angular momentum of the system is conserved. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \(\\{\mathrm{B}\\}\) 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 circular disc of radius \(\mathrm{R}\) is removed from a bigger disc of radius \(2 \mathrm{R}\). such that the circumferences of the disc coincide. The centre of mass of the remaining portion is \(\alpha R\) from the centre of mass of the bigger disc. The value of \(\alpha\) is. \(\\{\mathrm{A}\\} 1 / 2\) \\{B \\} \(1 / 6\) \\{C\\} \(1 / 4\) \(\\{\mathrm{D}\\}[(-1) / 3]\)

According to the theorem of parallel axis \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^{2}\) the graph between \(\mathrm{I} \rightarrow \mathrm{d}\) will be

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..... \(\\{\mathrm{A}\\}\left(\mathrm{MR}^{2} / 3\right)\) \(\\{B\\}\left(M R^{2} / 4\right)\) \(\\{\mathrm{C}\\}(3 / 8) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(3 / 2) \mathrm{MR}^{2}\)
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