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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

One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is... \(\\{\mathrm{A}\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(1 / 4) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(1 / 8) \mathrm{MR}^{2}\) \\{D \(\\} \sqrt{2} \mathrm{MR}^{2}\)

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

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