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Chapter 5: Problem 660

A circular disc of radius \(R\) is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of \(60 ?\) and released. Its angular velocity when it reaches the equilibrium position will be \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / 3 \mathrm{R})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\) \(\\{\mathrm{C}\\} \sqrt{(2 \mathrm{~g} / \mathrm{R})}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 \mathrm{~g} / \mathrm{R})}\)

Short Answer

Expert verified
The correct answer is the angular velocity ω = \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\), which is obtained by using the conservation of mechanical energy principle and solving for ω.

Step by step solution

01

Calculate the Initial Gravitational Potential Energy

First, let's calculate the initial gravitational potential energy when the disc is turned through an angle of 60°. The center of mass of the disc is located at a distance R/2 from the axis of rotation. In the initial position, the height of the center of mass above the equilibrium position can be calculated using trigonometry: Height = (R/2) * (1 - cos(60°)) Now, we can calculate the gravitational potential energy (PE_initial): PE_initial = m * g * Height, where m is the mass of the disc.
02

Calculate the Final Rotational Kinetic Energy

When the disc reaches the equilibrium position, its gravitational potential energy is converted to rotational kinetic energy (KE_final): KE_final = (1/2) * I * ω^2, where ω is the angular velocity, and I is the moment of inertia of the disc. The moment of inertia of the circular disc about the given axis can be calculated with the parallel axis theorem: I = (1/2) * m * R^2 + m * (R/2)^2 = (3/4) * m * R^2
03

Equate the Initial Gravitational Potential Energy to the Final Rotational Kinetic Energy

By conservation of energy, the initial gravitational potential energy equals the final rotational kinetic energy: PE_initial = KE_final m * g * Height = (1/2) * (3/4) * m * R^2 * ω^2
04

Solve for the Angular Velocity ω

Now, we can solve for the angular velocity ω: ω^2 = (2 * g * Height) / ((3/2) * R^2) Since Height = (R/2) * (1 - cos(60°)) = R/4, we can substitute it into the equation: ω^2 = (2 * g * (R/4)) / ((3/2) * R^2) ω^2 = (2 * g) / (3 * R) Now, we can find the angular velocity ω by taking the square root: ω = √((2 * g) / (3 * R)) So the correct answer is: \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\).

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Most popular questions from this chapter

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

A particle of mass \(\mathrm{m}\) slides down on inclined plane and reaches the bottom with linear velocity \(\mathrm{V}\). If the same mass is in the form of ring and rolls without slipping down the same inclined plane. Its velocity will be \(\\{\mathrm{A}\\} \mathrm{V}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~V})}\) \(\\{\mathrm{C}\\}(\mathrm{V} / \sqrt{2})\) \(\\{\mathrm{D}\\} 2 \mathrm{~V}\)

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 \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)

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 \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)

A constant torque of \(1500 \mathrm{Nm}\) turns a wheel of moment of inertia \(300 \mathrm{~kg} \mathrm{~m}^{2}\) about an axis passing through its centre the angular velocity of the wheel after 3 sec will be.......... $\mathrm{rad} / \mathrm{sec}$ \(\\{\mathrm{A}\\} 5\) \\{B \\} 10 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)
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