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Chapter 8: Problem 20

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.33 \mathrm{~s}\), the angular velocity of the wheel is \(4.96 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=23.0 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=46.0 \mathrm{~s}\) ?

Short Answer

Expert verified
The wheel rotates through an angle of approximately 1687 rad from t=0 to t=46.0 s.

Step by step solution

01

Calculate Angular Acceleration

Given that the wheel starts from rest, the initial angular velocity (w0) is 0 rad/s. The final angular velocity (w) at t = 2.33 s is 4.96 rad/s. We can use the equation \(w = w0 + \alpha*t\) where \(\alpha\) is the angular acceleration to solve for \(\alpha\). So, \(\alpha = (w - w0) / t = 4.96 / 2.33 = 2.13 \, rad/s^2\).
02

Determine Angle of Rotation for t=0 to t=23.0 s

Now, we find the angle of rotation for the first 23.0s. We know that the angular acceleration is constant during this time. Therefore, we can use the equation for angular displacement (\(\theta\)) for uniformly accelerated motion: \(\theta = w0*t + 0.5*\alpha*t^2\). Substituting the known values, we obtain \(\theta = 0.5*2.13*23.0^2 = 558.045 \, rad\).
03

Determine Angle of Rotation for t=23.0 s to t=46.0 s

For the period from 23.0 s to 46.0 s, we know that the wheel is not accelerating, so it's rotating at a uniform rate (the final angular velocity from the accelerated period). Hence, we can use the equation \(\theta = w*t\) to find the angle of rotation during this period, where w is the angular velocity at 23.0 s. The angular velocity is given by the final angular velocity from the accelerated period and \(w = w0 + \alpha*t = 0 + 2.13*23.0 = 48.99 \, rad/s\). So, \(\theta = 48.99*(46.0-23.0) = 1128.77 \, rad\).
04

Calculate Total Angle of Rotation from t=0 to t=46.0 s

Finally, add the angles of rotation calculated in steps 2 & 3 to find the total angle of rotation from t=0 to t=46.0 s. Hence, the total angle of rotation \(\theta_total = 558.045 + 1128.77 = 1686.815 \, rad\).

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Most popular questions from this chapter

A speedometer on the front wheel of a bicycle gives a reading that is directly proportional to the angular speed of the wheel. Suppose that such a speedometer is calibrated for a wheel of diameter \(72 \mathrm{~cm}\) but is mistakenly used on a wheel of diameter \(62 \mathrm{~cm} .\) Would the linear speed reading be wrong? If so, in what sense and by what fraction of the true speed?

The angular speed of an automobile engine is increased uniformly from 1170 rev/min to 2880 rev/min in \(12.6 \mathrm{~s}\). (a) Find the angular acceleration in rev/min \(^{2} .(b)\) How many revolutions does the engine make during this time?

An object moves in the \(x y\) plane such that \(x=R \cos \omega t\) and \(y=R \sin \omega t\). Here \(x\) and \(y\) are the coordinates of the object, is the time, and \(R\) and \(\omega\) are constants. (a) Eliminate \(t\) between these equations to find the equation of the curve in which the object moves. What is this curve? What is the meaning of the constant \(\omega ?(b)\) Differentiate the equations for \(x\) and \(y\) with respect to the time to find the \(x\) and \(y\) components of the velocity of the body, \(v_{x}\) and \(v_{y}\). Combine \(v_{x}\) and \(v_{y}\) to find the magnitude and direction of \(v .\) Describe the motion of the object. ( \(c\) ) Differentiate \(v_{x}\) and \(v_{y}\) with respect to the time to obtain the magnitude and direction of the resultant acceleration.

A planet \(P\) revolves around the Sun in a circular orbit, with the Sun at the center, which is coplanar with and concentric to the circular orbit of Earth \(E\) around the Sun. \(P\) and \(E\) revolve in the same direction. The times required for the revolution of \(P\) and \(E\) around the Sun are \(T_{P}\) and \(T_{E}\). Let \(T_{S}\) be the time required for \(P\) to make one revolution around the Sun relative to \(E\) : show that \(1 / T_{S}=1 / T_{E}-1 / T_{P}\). Assume \(T_{P}>T_{E}\).

Our Sun is \(2.3 \times 10^{4}\) ly (light-years) from the center of our Milky Way galaxy and is moving in a circle around this center at a speed of \(250 \mathrm{~km} / \mathrm{s}\). (a) How long does it take the Sun to make one revolution about the galactic center? \((b)\) How many revolutions has the Sun completed since it was formed about \(4.5 \times 10^{9}\) years ago?
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