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Chapter 5: Problem 605

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..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)

Short Answer

Expert verified
The new angular velocity is \(\omega' = \frac{M\omega}{M + m}\), which corresponds to option C.

Step by step solution

01

Calculate the initial angular momentum

Before the wax pieces are added, the gramophone record's angular momentum can be found using the formula: \[L_{i} = I_{record} \omega\] Where \(L_{i}\) is the initial angular momentum, \(I_{record}\) is the moment of inertia of the record, and \(\omega\) is the given angular speed. For a disc, the moment of inertia can be calculated as: \[I_{record} = \frac{1}{2} MR^2\] Now, substitute this expression into the initial angular momentum equation: \[L_{i} = \frac{1}{2} MR^2 \omega\]
02

Determine the final angular momentum

After placing the wax pieces on the record, the moment of inertia of each wax piece can be calculated as: \[I_{wax} = mr^2\] Where \(m\) is the mass of each wax piece, and \(r = \frac{R}{2}\) is the distance from the center of the record. The total moment of inertia for the new system is the sum of the moments of inertia of the record and both of the wax pieces, so we can write: \[I_{total} = I_{record} + 2I_{wax}\] Substitute the expressions for the moments of inertia into this equation: \[I_{total} = \frac{1}{2} MR^2 + 2m\left(\frac{R}{2}\right)^2\] The final angular momentum of the system is: \[L_{f} = I_{total} \omega'\] Where \(\omega'\) is the new angular velocity.
03

Set up the conservation of angular momentum equation

As no external torques are acting on the system, we can equate the initial and final angular momentum of the system: \[L_{i} = L_{f}\] Now, substitute the expressions for the initial and final angular momentum into this equation: \[\frac{1}{2} MR^2 \omega = I_{total} \omega'\]
04

Solve for the new angular velocity

To find the new angular velocity \(\omega'\), we can divide both sides of the conservation of angular momentum equation by \(I_{total}\): \[\omega' = \frac{\frac{1}{2} MR^2 \omega}{\frac{1}{2} MR^2 + 2m\left(\frac{R}{2}\right)^2}\] Simplify the expression: \[\omega' = \frac{MR^2\omega}{MR^2 + \frac{1}{2} mR^2}\] Finally, we can see that the correct answer is: \[\omega' = \frac{M\omega}{M + m}\] Which corresponds to option C.

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Most popular questions from this chapter

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 \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

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 spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)

A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be.... \(\\{\mathrm{A}\\} 15: 14\) \(\\{\mathrm{B}\\} 14: 15\) \(\\{\mathrm{C}\\} 5: 3\) \(\\{\mathrm{D}\\} 3: 5\)

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) 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 end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)
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