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Chapter 5: Problem 557

A wheel rotates with a constant acceleration of $2.0 \mathrm{rad} / \mathrm{sec}^{2}$ If the wheel start from rest. The number of revolution it makes in the first ten seconds will be approximately. \(\\{\mathrm{A}\\} 8\) \\{B \\} 16 \(\\{\mathrm{C}\\} 24\) \(\\{\mathrm{D}\\} 32\)

Short Answer

Expert verified
The number of revolutions made by the wheel in the first 10 seconds is approximately \(\\{B\\} 16\).

Step by step solution

01

Write down relevant equations of motion for constant angular acceleration

There are two main equations that we can use for this problem: 1. Angular displacement equation (final angle, initial angle, initial angular velocity, time, and angular acceleration): \(\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\) 2. Angular velocity equation (final angular velocity, initial angular velocity, angular acceleration, and time): \(\omega = \omega_0 + \alpha t\) In this problem, we know that the initial angular velocity \(\omega_0 = 0\), the angular acceleration \(\alpha = 2.0 \, \text{rad/sec}^2\), and the time \(t = 10 \, \text{sec}\). We need to find the angular displacement in radians, \(\theta\), to calculate the number of revolutions.
02

Calculate the angular displacement using the first equation of motion

Using the equation \(\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\), we can substitute our known values and calculate the angular displacement: \(\theta = 0 + 0 \cdot 10 + \frac{1}{2}(2.0) \cdot 10^2\) Calculating this value, we get: \(\theta = 100 \, \text{rad}\)
03

Convert angular displacement in radians to revolutions

Since there are 2π radians in one revolution, we can divide the angular displacement by 2π to get the number of revolutions: Number of revolutions = \(\frac{100 \, \text{rad}}{2π}\) Approximating π as 3.14, we have: Number of revolutions = \(\frac{100}{6.28} \approx 15.9\)
04

Choose the closest answer from the options

From our calculations, the number of revolutions is approximately 15.9. Looking at the answer choices, option B (16) is the closest to our calculated value. Therefore, the answer to this problem is: \(\\{B\\} 16\)

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Most popular questions from this chapter

In a bicycle the radius of rear wheel is twice the radius of front wheel. If \(\mathrm{r}_{\mathrm{F}}\) and \(\mathrm{r}_{\mathrm{r}}\) are the radius, \(\mathrm{v}_{\mathrm{F}}\) and \(\mathrm{v}_{\mathrm{r}}\) are speed of top most points of wheel respectively then... \(\\{\mathrm{A}\\} \mathrm{v}_{\mathrm{r}}=2 \mathrm{v}_{\mathrm{F}}\) $\\{\mathrm{B}\\} \mathrm{v}_{\mathrm{F}}=2 \mathrm{v}_{\mathrm{r}} \quad\\{\mathrm{C}\\} \mathrm{v}_{\mathrm{F}}=\mathrm{v}_{\mathrm{r}}$ \(\\{\mathrm{D}\\} \mathrm{v}_{\mathrm{F}}>\mathrm{v}_{\mathrm{r}}\)

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)

A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm} .\) is removed from tve \(\mathrm{x}\) edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate. $\\{\mathrm{A}\\}-7 \mathrm{~cm} \quad\\{\mathrm{~B}\\}+9 \mathrm{~cm} \quad\\{\mathrm{C}\\}-9 \mathrm{~cm} \quad\\{\mathrm{D}\\}+7 \mathrm{~cm}$

Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)

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. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)
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