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A wheel of diameter \(40.0 \mathrm{~cm}\) starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) \(a_{\mathrm{rad}}=v^{2} / r\)

Short Answer

Expert verified
The radial acceleration of a point on the rim when the wheel completes its second revolution is \(4.8\pi\) m/s², computed using both formulas.

Step by step solution

01

Calculate the angular displacement

To find the angular velocity (\(\omega\)) at the point where the wheel completes its second revolution, we first need to calculate the angular displacement.Two revolutions equal to \(2(2\pi) = 4\pi\) rad, since one complete revolution equals \(2\pi\) rad.
02

Calculate the angular velocity (\(\omega\))

The formula for angular velocity when initial angular velocity (\(\omega_0\)) is zero, angular displacement is \(\theta\) & angular acceleration is \(\alpha\) is \(\omega = \sqrt{2\alpha\theta}\). Substituting the provided values: \(\omega = \sqrt{2(3.0)4\pi} = \sqrt{24 \pi}\) rad/s.
03

Calculate radial acceleration using \(a_{\mathrm{rad}}=\omega^{2} r\)

To calculate the radial (centripetal) acceleration using the first relationship, we substitute the previously calculated \(\omega\) and provided radius (which is half of the diameter, \(r=0.2\) m) into the formula \(a_{\mathrm{rad}}=\omega^{2} r\). Hence, \(a_{\mathrm{rad1}}= (\sqrt{24 \pi})^2 (0.2) = 4.8\pi\) m/s².
04

Calculate the linear velocity (v)

The formula to calculate the linear velocity when we have the angular velocity is \(v = \omega r\). Substituting the previously calculated \(\omega\) and \(r=0.2\) m we get \(v = \sqrt{24 \pi}(0.2)\) m/s.
05

Calculate radial acceleration using \(a_{\mathrm{rad}}=v^{2} / r\)

Substituting the calculated linear velocity \(v\) and radius \(r\) into the second formula to compute radial acceleration, we get \(a_{\mathrm{rad2}}=(\sqrt{24 \pi}(0.2))^2 / 0.2 = 4.8\pi\) m/s². Both \(a_{\mathrm{rad1}}\) and \(a_{\mathrm{rad2}}\) are consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Acceleration
Centripetal acceleration is a key concept when discussing objects in circular motion. It refers to the acceleration that keeps an object moving along a circular path. This acceleration is directed towards the center of the circle. It is essential for maintaining the circular motion as it counteracts the inertia of the object, which would otherwise move in a straight line.

Centripetal acceleration, denoted as \(a_{\mathrm{rad}}\), can be calculated using two primary relationships. The first formula is \(a_{\mathrm{rad}} = \omega^{2} r\), where \(\omega\) is the angular velocity and \(r\) is the radius of the circular path. This formula is helpful when the angular velocity is known.
  • For example, if a wheel rotates with a certain angular velocity, substituting \(\omega\) and \(r\) gives the centripetal acceleration at the rim of the wheel.
  • The second formula is \(a_{\mathrm{rad}} = \frac{v^{2}}{r}\). Here, \(v\) represents the linear velocity. This formula is useful when you know the speed of the object along the circular path.
Both formulas ultimately give you the centripetal acceleration, which is crucial for understanding the dynamics of circular motion.
Angular Velocity
Angular velocity, represented by \(\omega\), describes how quickly an object rotates around a particular axis. It is a vector quantity, with both direction and magnitude, and is measured in radians per second (rad/s).

To calculate angular velocity, especially from a state of rest, you can use the formula \(\omega = \sqrt{2\alpha\theta}\), where \(\alpha\) stands for angular acceleration and \(\theta\) is the angular displacement. This calculation is crucial when dealing with objects that experience constant angular acceleration, like a spinning wheel.
  • For instance, when a wheel completes its second revolution, you can determine its angular velocity by computing its angular displacement (using total revolutions) and applying the above formula.
  • Remember, knowing angular velocity is critical, as it links to calculating other quantities such as linear velocity and centripetal acceleration.
Angular velocity gives insight into how fast an object spins, which impacts various physical situations, particularly those involving rotational dynamics.
Angular Displacement
Angular displacement, symbolized by \(\theta\), is a measure of the angle through which an object rotates about a particular axis. Unlike linear displacement, which measures distance along a straight path, angular displacement measures that distance in terms of the spread angle, typically in radians.

To find angular displacement for a rotating object, you multiply the number of revolutions by \(2\pi\), since one complete revolution corresponds to \(2\pi\) radians. For example, if a wheel completes two revolutions, the angular displacement is \(4\pi\) radians.
  • Angular displacement is integral in calculating other rotational motion-related variables, including angular velocity and angular acceleration.
  • It gives a clear picture of the "distance" an object has covered on its rotational path without specifying the path's length.
Understanding angular displacement is foundational in studying rotational dynamics, and by understanding changes in angular displacement, you get insights into overall rotation dynamics.

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Most popular questions from this chapter

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius \(12.0 \mathrm{~cm} .\) If the angular speed of the front sprocket is 0.600 rev \(/ \mathrm{s},\) what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be \(5.00 \mathrm{~m} / \mathrm{s} ?\) The rear wheel has radius \(0.330 \mathrm{~m}\).

An advertisement claims that a centrifuge takes up only \(0.127 \mathrm{~m}\) of bench space but can produce a radial acceleration of \(3000 g\) at 5000 rev \(/\) min. Calculate the required radius of the centrifuge. Is the claim realistic?

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) where \(\quad \gamma=5.00 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.800 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{~s}\) and the average angular acceleration \(\alpha_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{~s}\). How do these two quantities compare? If they are different, why?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)

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