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Chapter 5: Problem 559

A wheel of mass \(10 \mathrm{~kg}\) has a moment of inertia of $160 \mathrm{~kg} \mathrm{~m}$ radius of gyration will be $\begin{array}{llll}\\{\mathrm{A}\\} 10 & \\{\mathrm{~B}\\} 8 & \\{\mathrm{C}\\} 6 & \\{\mathrm{D}\\} 4\end{array}$

Short Answer

Expert verified
The radius of gyration, \(k\), can be found using the formula \( k = \sqrt{\frac{I}{m}} \), where \(I\) is the moment of inertia and \(m\) is the mass of the object. Substituting the given values (\(m = 10 \, kg\) and \(I = 160 \, kg \cdot m^{2}\)), we get \( k = \sqrt{\frac{160 \, kg \cdot m^{2}}{10 \, kg}} \). Simplifying the equation, we get \( k = 4 \, m \). Hence, the radius of gyration of the wheel is 4 m. The correct answer is 'D'.

Step by step solution

01

Understand the formula

The formula to find the radius of gyration, \( k \), is given by \( k = \sqrt{\frac{I}{m}} \), where \( I \) is the moment of inertia and \( m \) is the mass of the object. This formula allows us to calculate the hypothetical distance at which the entire mass can be assumed to be concentrated without changing the moment of inertia.
02

Substitute in the given values

In this case, we are given that mass \( m = 10 \, kg \) and moment of inertia \( I = 160 \, kg \cdot m^{2} \). Substitute these values into the formula to get \( k = \sqrt{\frac{160 \, kg \cdot m^{2}}{10 \, kg}} \).
03

Simplify the equation

Our equation \( k = \sqrt{\frac{160 \, kg \cdot m^{2}}{10 \, kg}} \) can now be simplified. The \( kg \) units on the top and bottom cancel out and dividing 160 by 10 gives us 16, so the equation becomes \( k = \sqrt{16 \, m^{2}} \).
04

Solve the equation

The square root of 16 is 4, so our final answer is \( k =4 \, m \). Thus, the radius of gyration of the wheel is 4 m. Therefore, the answer to the exercise is 'D'.

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Most popular questions from this chapter

Two spheres each of mass \(\mathrm{M}\) and radius \(\mathrm{R} / 2\) are connected with a mass less rod of length \(2 \mathrm{R}\) as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? \(\\{\mathrm{A}\\}(21 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(2 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(5 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(5 / 21) \mathrm{MR}^{2}\)

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 \(\\} .400 \pi \mathrm{Nm}\) \(\\{\mathrm{D}\\} 300 \pi \mathrm{Nm}\)

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) \(4 \mathrm{~cm}\)

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first two second it rotate through an angle \(\theta_{1}\), in the next \(2 \mathrm{sec}\). it rotates through an angle \(\theta_{2}\), find the ratio \(\left(\theta_{2} / \theta_{1}\right)=\) \(\\{\mathrm{A}\\} 1\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} 3\) \(\\{\mathrm{D}\\} 4\)

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 \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)
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