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Chapter 5: Problem 613

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

Short Answer

Expert verified
The moment of inertia of the meter scale about the given axis is approximately \(0.148\,\text{kgm}^2\).

Step by step solution

01

Understand the given information

We are given: Mass of the meter scale (m) = 0.6 kg Position of the axis from one end (x) = 30 cm = 0.3 meters (since 1 meter = 100 cm) Total length of the scale (L) = 1 meter
02

Moment of inertia formula for the axis at one end

The moment of inertia of a rod of mass(m) and length (L) about an axis perpendicular to the rod and passing through one end is given by the formula: \(I_{end}=\frac{1}{3}mL^2\)
03

Calculate the moment of inertia at one end

Now, let's calculate the moment of inertia at one end using the given values: \(I_{end} = \frac{1}{3}(0.6\,\text{kg})(1\,\text{m})^2 =\frac{1}{3}(0.6)= 0.2\,\text{kgm}^2\)
04

Moment of inertia formula for a perpendicular axis

Now, we need to find the moment of inertia of the rod about the given axis at 30 cm position, For this, we will use the perpendicular axis theorem: \(I_x = I_{end} - mx^2\) Where, \(I_x\) = Moment of inertia about the given axis \(I_{end}\) = Moment of inertia about the axis at one end m = mass of the rod x = distance of the given axis from one end
05

Calculate the moment of inertia at the given axis

Now, let's find the moment of inertia at the given axis using the perpendicular axis theorem: \(I_x = 0.2\,\text{kgm}^2 - (0.6\,\text{kg})(0.3\,\text{m})^2 = 0.2 - (0.6)(0.09) = 0.2 - 0.054 = 0.146\,\text{kgm}^2\) Comparing the calculated value with the options given, we see that the answer is approximately equal to option D: \(\{\text{D}\} 0.148 \,\text{kgm}^2\) Therefore, the moment of inertia of the meter scale about the given axis is approximately \(0.148\,\text{kgm}^2\).

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Most popular questions from this chapter

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)

Two point masses of \(0.3 \mathrm{~kg}\) and \(0.7 \mathrm{~kg}\) are fixed at the ends of a rod of length \(1.4 \mathrm{~m}\) and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of ..... \(\\{\mathrm{A}\\} 0.4 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{B \(0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{C\\} \(0.7 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\) \\{D \(\\} 0.98 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\)

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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