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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

A uniform disc of mass \(500 \mathrm{~kg}\) and radius \(2 \mathrm{~m}\) is rotating at the rate of \(600 \mathrm{r}\).p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of \(100 \mathrm{sec}\) ? \(\\{\mathrm{A}\\} 600 \pi \mathrm{Nm}\) \\{B \(\\} 500 \pi \mathrm{Nm}\) \\{C \(\\} .400 \pi \mathrm{Nm}\) \(\\{\mathrm{D}\\} 300 \pi \mathrm{Nm}\)

From a circular disc of radius \(\mathrm{R}\) and mass \(9 \mathrm{M}\), a small disc of radius \(\mathrm{R} / 3\) is removed from the disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through \(\mathrm{O}\) is.... \(\\{\mathrm{A}\\} 4 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(40 / 9) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(37 / 9) \mathrm{MR}^{2}\)

Two point masses of \(0.3 \mathrm{~kg}\) and \(0.7 \mathrm{~kg}\) are fixed at the ends of a rod of length \(1.4 \mathrm{~m}\) and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of ..... \(\\{\mathrm{A}\\} 0.4 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{B \(0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{C\\} \(0.7 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\) \\{D \(\\} 0.98 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\)

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\\} \(1 / 4\) \(\\{\mathrm{D}\\}[(-1) / 3]\)

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 acceleration of the centre of mass of the disc is \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)
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