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Chapter 8: Q60P (page 198)

Question:(I) What is the angular momentum of a 0.270-kg ball revolving on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.4 rad/s?

Short Answer

Expert verified

The angular momentum of the ball is \(5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

Step by step solution

01

Given data

The mass of the ball is\(m = 0.270\;{\rm{kg}}\).

The radius of the circle is\(r = 1.35\;{\rm{m}}\).

The angular speed is \(\omega = 10.4\;{\rm{rad/s}}\).

02

Understanding angular momentum

In this problem, use the product of the moment of inertia of the ball and its angular speed to evaluate the angular momentum.

03

Determine the angular momentum of the ball

The relation of angular momentum can be written as:

\(\begin{aligned}{l}L &= I\omega \\L &= \left( {m{r^2}} \right)\omega \end{aligned}\)

Here, Iis the moment of inertia of the ball.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}L &= \left( {0.270\;{\rm{kg}} \times {{\left( {1.35\;{\rm{m}}} \right)}^2}} \right)\left( {10.4\;{\rm{rad/s}}} \right)\\L &= 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

Thus, \(L = 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\) is the required angular momentum.

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Most popular questions from this chapter

The bolts on the cylinder head of an engine require tightening to a torque of 95 m N. If a wrench is 28 cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15 mm across (Fig. 8–44), estimate the force applied near each of the six points by a wrench.

Assume that a 1.00-kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, as shown in Fig. 8–46. The ball is accelerated uniformly from rest to 8.5 m/s in 0.38 s, at which point it is released. Calculate (a) the angular acceleration of the arm and (b) the force required for the triceps muscle. Assume that the forearm has a mass of 3.7 kg, and it rotates like a uniform rod about an axis at its end.

FIGURE 8-46

Problems 35 and 36

(II) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (a) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. (Assume the Earth is a uniform sphere with\({\bf{mass}} = {\bf{6}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{24}}}}{\bf{ kg}}\),\({\bf{radius}} = {\bf{6}}{\bf{.4 \times 1}}{{\bf{0}}^{\bf{6}}}{\bf{ m}}\)and is\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{8}}}{\bf{ km}}\)from the Sun.)

I) Water drives a waterwheel (or turbine) of radius R = 3.0 m as shown in Fig. 8–66. The water enters at a speed\({v_1} = 7.0\;{\rm{m/s}}\)and exits from the waterwheel at a speed\({v_2} = 3.8\;{\rm{m/s}}\). (a) If 85 kg of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? (b) What is the torque the water applies to the waterwheel? (c) If the water causes the waterwheel to make one revolution every 5.5 s, how much power is delivered to the wheel?

An Atwood machineconsists of two masses,\({m_A} = {\bf{65 kg}}\) and\({m_B} = {\bf{75 kg}}\) connected by a massless inelastic cord that passes over a pulley free to rotate, Fig. 8 52. The pulley is a solid cylinder of radius\(R = {\bf{0}}{\bf{.45 m}}\) and mass 6.0 kg. (a) Determine the acceleration of each mass. (b) What % error would be made if the moment of inertia of the pulley is ignored? (Hint: The tensions\({F_{TA}}\) and\({F_{TB}}\)are not equal. We discussed the Atwood machine in Example 4–13, assuming I = 0 for the pulley.)

FIGURE 8-52 Problem 47.Atwood machine.
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