The crane shown in the figure is lifting a 226-kg crate upward with an acceleration of 4.52 m/s². The cable from the crate passes over asolid cylindrical pulley at the top of the boom. The pulley has a mass of 148 kg. The cable is then wound onto a hollow cylindrical drumthat is mounted on the deck of the crane. The mass of the drum is 267 kg, and its radius (the same as that of the pulley) is 0.237 m. Theengine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore themass of the cable.Number iUnitsDrumEFFIPulley-Boom

The crane shown in the figure is lifting a 226-kg crate upward with an acceleration of 4.52 m/s². The cable from the crate pa solid cylindrical pulley at the top of the boom. The pulley has a mass of 148 kg. The cable is then wound onto a hollow cylind that is mounted on the deck of the crane. The mass of the drum is 267 kg, and its radius (the same as that of the pulley) is 0.2 engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? mass of the cable. Pulley-

The crane shown in the figure is lifting a 226-kg crate upward with an acceleration of 4.52 m/s². The cable from the crate pa solid cylindrical pulley at the top of the boom. The pulley has a mass of 148 kg. The cable is then wound onto a hollow cylind that is mounted on the deck of the crane. The mass of the drum is 267 kg, and its radius (the same as that of the pulley) is 0.2 engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? mass of the cable. Pulley-

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter11: Angular Momentum

Section: Chapter Questions

Problem 49P: The blades of a wind turbine are 30 m in length and rotate at a maximum rotation rate of 20 rev/min....

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CHECK and THINK Our results give us a way to think about how a person might steer a unicycle. Consider the person, the unicycle, and the Earth as the system. No net torque acts on the system. Ignoring the motion of the Earth, Figure 13.37A shows the initial angular momentum of the system with the unicycle in motion: Li=Ltire. The person leans to his left so that the angular momentum of the tire rotates downward. The total angular momentum must still point to the right (Fig. 13.37B), so the angular momentum of the person must be upward to compensate. The persons angular velocity therefore points upward (parallel to his own angular momentum), and he is able to make a turn. Another way to analyze this situation is to exclude the Earth from the system and calculate the torque done by gravity. This approach is left as a homework problem. FIGURE 13.37 C Reanalyze the unicycles motion in Example 13.15 (page 382). This time, leave the Earth out of the system and explain how the torque exerted by gravity causes the unicycle to turn. Your explanation should involve a diagram.
Check Your Understanding What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below? A uniform thin disk about an axis through the center Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center (Figure 10.27). Figure 10.27 Calculating the moment of inertia for a thin disk about an axis through its center.