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Chapter 5: Problem 651

A circular disc of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is \(\\{\mathrm{A}\\}(1 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{B}\\}(1 / 2) \mathrm{mv}^{2}\) \(\\{\mathrm{C}\\}(3 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{D}\\} \mathrm{mv}^{2}\)

Short Answer

Expert verified
The total kinetic energy of the circular disc is \(\frac{3}{4}mv^2\). Hence, the correct answer is \(\boxed{\text{C} (3 / 4) \mathrm{mv}^{2}}\).

Step by step solution

01

Identify the given variables.

In this problem, we are given the following variables: 1. Mass of the circular disc (m) 2. Radius of the circular disc (r) 3. Constant speed of the circular disc (v)
02

Calculate the translational kinetic energy.

The translational kinetic energy of the disc, moving with a constant speed v, is given by the formula: Translational Kinetic Energy = \( \frac{1}{2} * m * v^2 \)
03

Calculate the rotational kinetic energy.

For a rolling object, the moment of inertia (I) for a circular disc about its center of mass is given by the formula: Moment of Inertia (I) = \( \frac{1}{2} * m * r^2 \) The rotational kinetic energy is given by the formula: Rotational Kinetic Energy = \( \frac{1}{2} * I * ω^2 \) Since we are given the constant speed (v) and we know that the object is rolling without slipping, we can find the angular velocity (ω) using the formula: \( ω = \frac{v}{r} \) Now we plug the values into the rotational kinetic energy formula: Rotational Kinetic Energy = \( \frac{1}{2} * (\frac{1}{2} * m * r^2) * (\frac{v}{r})^2 \) Simplify the equation: Rotational Kinetic Energy = \( \frac{1}{4} * m * v^2 \)
04

Calculate the total kinetic energy.

Now we can find the total kinetic energy by summing both the translational and rotational kinetic energies: Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy Total Kinetic Energy = \( (\frac{1}{2} * m * v^2) + (\frac{1}{4} * m * v^2) \) Total Kinetic Energy = \( (\frac{3}{4} * m * v^2) \)
05

Find the correct answer choice.

Now we have the total kinetic energy of the circular disc as \( (\frac{3}{4} * m * v^2) \). Looking at the given options, the correct answer is: \(\boxed{\text{C} (3 / 4) \mathrm{mv}^{2}}\)

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Most popular questions from this chapter

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

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. If the disc is replaced by a ring of the same mass \(\mathrm{M}\) and the same radius \(R\), the ratio of the frictional force on the ring to that on the disc will be \(\\{\mathrm{A}\\} 3 / 2\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} \sqrt{2}\) \(\\{\mathrm{D}\\} 1\)

The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is \(\mathrm{I}_{1}\). The same rod is bent in to a ring and its moment of inertia about a diameter is \(\mathrm{I}_{2}\), Then \(\left[\mathrm{I}_{1} / \mathrm{I}_{2}\right]\) is. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

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