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Chapter 21: Problem 31

A cylinder placed in frictionless bearings is set rotating about its axis. The cylinder is then heated, without mechanical contact, until its radius is increased by \(0.18 \%\). What is the percent change in the cylinder's ( \(a\) ) angular momentum, \((b)\) angular velocity, and ( \(c\) ) rotational energy?

Short Answer

Expert verified
The percent change in the cylinder's angular momentum is 0%, in angular velocity is -0.36%, and in rotational energy is -0.72%.

Step by step solution

01

Determine the initial moment of inertia, angular momentum and rotational energy

Initially, let the energy, angular momentum and moment of inertia be E, L, and I respectively with the radius r. Thus, \( I = 0.5 m r^2 \), the angular momentum \( L = I w = 0.5 m r^2 w \), and the rotational energy \( E = 0.5 I w^2 = 0.25 m r^2 w^2 \). Note: All quantities are initial quantities.
02

Determine the final moment of inertia, angular momentum and rotational energy

After heating, the radius increases by 0.18%, which means the new radius r' is \( 1.0018 r \). The moment of inertia is now \( I' = 0.5 m (r')^2 = 0.5 m (1.0018 r)^2 \). Since there are no external torques, the angular momentum must be conserved, so we have \( L = I' w' \) where \( w' \) is the new angular velocity and rearrange to find \( w' = L / I' = 0.5 m r^2 w / 0.5 m (1.0018 r)^2 = w / 1.0018^2 \). The new rotational energy is \( E' = 0.5 I' (w')^2 = 0.25 m (1.0018 r)^2 (w / 1.0018^2)^2 \).
03

Compute the percentage changes

The percentage change in angular momentum is \((L' - L) / L * 100% = (L - L) / L * 100% = 0% \) since angular momentum is conserved. The percentage change in angular velocity is \((w' - w) / w * 100% = (w / 1.0018^2 - w) / w * 100% = -0.36% \). The percentage change in rotational energy is \( (E' - E) / E * 100% = (0.25 m (1.0018 r)^2 (w / 1.0018^2)^2 - 0.25 m r^2 w^2) / 0.25 m r^2 w^2 * 100% = -0.72% \).

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Most popular questions from this chapter

An automobile tire has a volume of \(988 \mathrm{in} .^{3}\) and contains air at a gauge pressure of \(24.2 \mathrm{lb} / \mathrm{in}^{2}\) where the temperature is \(-2.60^{\circ} \mathrm{C}\). Find the gauge pressure of the air in the tire when its temperature rises to \(25.6^{\circ} \mathrm{C}\) and its volume increases to \(1020 \mathrm{in} .{ }^{3}\). (Hint: It is not necessary to convert from British to SI units. Why? Use \(p_{\text {atm }}=14.7 \mathrm{lb} / \mathrm{in}^{2} .\) )

A quantity of ideal gas at \(12.0^{\circ} \mathrm{C}\) and a pressure of \(108 \mathrm{kPa}\) occupies a volume of \(2.47 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? \((b)\) If the pressure is now raised to \(316 \mathrm{kPa}\) and the temperature is raised to \(31.0^{\circ} \mathrm{C}\), how much volume will the gas now occupy? Assume there are no leaks.

An air bubble of \(19.4 \mathrm{~cm}^{3}\) volume is at the bottom of a lake \(41.5 \mathrm{~m}\) deep where the temperature is \(3.80^{\circ} \mathrm{C}\). The bubble rises to the surface, which is at a temperature of \(22.6^{\circ} \mathrm{C}\). Take the temperature of the bubble to be the same as that of the surrounding water and find its volume just before it reaches the surface.

( \(a\) ) Prove that the change in period \(P\) of a physical pendulum with temperature is given by \(\Delta P=\frac{1}{2} \alpha P \Delta T .(b)\) A clock pendulum made of invar has a period of \(0.500 \mathrm{~s}\) and is accurate at \(20^{\circ} \mathrm{C}\). If the clock is used in a climate where the temperature averages \(30^{\circ} \mathrm{C}\), what approximate correction to the time given by the clock is necessary at the end of 30 days?

(a) Prove that the change in rotational inertia \(I\) with temperature of a solid object is given by \(\Delta I=2 \alpha I \Delta T .(b)\) A thin uniform brass rod, spinning freely at 230 rev/s about an axis perpendicular to it at its center, is heated without mechanical contact until its temperature increases by \(170 \mathrm{C}^{\circ}\). Calculate the change in angular velocity.
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