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Chapter 5: Problem 600

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight \(\mathrm{mg}\) is attached to the end of the cord and falls from the rest. After falling through the distance \(\mathrm{h}\). the angular velocity of the wheel will be.... \(\\{B\\}\left[2 m g h /\left(I+m r^{2}\right)\right]\) $\\{\mathrm{C}\\}\left[2 \mathrm{mgh} /\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}$ \(\\{\mathrm{D}\\} \sqrt{(2 \mathrm{gh})}\)

Short Answer

Expert verified
The angular velocity of the wheel after the weight falls from rest is given by \(\omega = \sqrt{\frac{2mgh}{mr^2 + I}}\), which corresponds to option \(\boxed{C\left[2 \mathrm{mgh}/\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}}\).

Step by step solution

01

Calculate the initial mechanical energy

First, we need to find the initial mechanical energy, which is the potential energy of the weight. The potential energy can be calculated as: \[U_{initial} = mgh\]
02

Calculate the final mechanical energy

When the weight falls, the wheel gains kinetic energy from the weight's potential energy. The kinetic energy will be split into two parts: linear kinetic energy of the weight and rotational kinetic energy of the wheel. At the final point, the potential energy will be zero. Linear kinetic energy of the weight: \[K_{linear} = \frac{1}{2}mv^2\] Rotational kinetic energy of the wheel: \[K_{rotational} = \frac{1}{2}I\omega^2\] Total kinetic energy (as potential energy has been converted to kinetic energy): \[U_{final} = K_{linear} + K_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\]
03

Conserve mechanical energy

The sum of all initial energies should equal the sum of all final energies. Thus, we have the following equation: \[U_{initial} = U_{final}\] \[mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\]
04

Relate the linear velocity of the weight to the angular velocity

The linear velocity of the weight can be related to the angular velocity of the wheel by the following equation: \[v = r\omega\]
05

Substitute linear velocity in the conservation of energy equation

We substitute the linear velocity with the angular velocity equation from step 4 to get: \[mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}I\omega^2\]
06

Simplify and solve for the angular velocity

Now, we need to simplify and solve for the angular velocity of the wheel: \[mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}I\omega^2\] \[2mgh = (mr^2 + I)\omega^2\] Finally, solving for \(\omega\), we get: \[\omega = \sqrt{\frac{2mgh}{mr^2 + I}}\] This matches option C: \[\boxed{C\left[2 \mathrm{mgh}/\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}}\]

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Most popular questions from this chapter

Particles of \(1 \mathrm{gm}, 1 \mathrm{gm}, 2 \mathrm{gm}, 2 \mathrm{gm}\) are placed at the corners \(A, B, C, D\), respectively of a square of side $6 \mathrm{~cm}$ as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. (A) \(1 \mathrm{~cm}\) (B) \(2 \mathrm{~cm}\) (C) \(3 \mathrm{~cm}\) (D) \(4 \mathrm{~cm}\)

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Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius $6 \mathrm{~cm}\(. If the distance between their centres is \)3.2 \mathrm{~cm}$, what is the shift in the centre of mass of the disc... $\begin{array}{llll}\\{\mathrm{A}\\}-0.4 \mathrm{~cm} & \\{\mathrm{~B}\\}-2.4 \mathrm{~cm} & \\{\mathrm{C}\\}-1.8 \mathrm{~cm} & \\{\mathrm{D}\\} & 1.2 \mathrm{~cm}\end{array}$
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