Explanation Given information: The mass ( m ) of the rotor is 25 kg. The time ( t ) is required for the rotor to coast to rest is 4.2 min. The angular velocity ( ω 1 ) of rotor is 3,600 rpm. The kinetic friction ( M ) produces a couple of magnitude 1.2 N ⋅ m Calculation: Consider principle of impulse and momentum. Syst Momenta 1 + Syst Ext Imp 1 → 2 = Syst Momenta 2 Sketch the impulse and moment diagram as shown in Figure (1). Refer Figure (1), Take moment about axle A . Σ M A = 0 I ¯ ω 1 − M t = 0 I ¯ = M t ω 1 Substitute 1.2 N ⋅ m for M , 4.2 min for t , and 3,600 rpm for ω 1 . I ¯ = ( 1.2 ) ( 4

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Free Damped Vibrations

Free vibration is a type of vibration in which the external force is removed after giving the initial displacement to a body/system. In other words, in this type of vibration, force is applied at the initial displacement of the system, and after the init…

Question

Chapter 17.2, Problem 17.52P

To determine

Find the centroidal radius of gyration for the rotor.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 17.2, Problem 17.3IMD

Chapter 17.2, Problem 17.53P

Students have asked these similar questions

An AISI 1018 steel ball with 1.100-in diameter is used as a roller between a flat plate made from 2024 T3 aluminum and a flat table surface made from ASTM No. 30 gray cast iron. Determine the maximum amount of weight that can be stacked on the aluminum plate without exceeding a maximum shear stress of 19.00 kpsi in any of the three pieces. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pmax 0.4 90 0.6 στ Tmax 0.2 0.5a a 1.5a 2a 2.5a За Distance from contact surface The maximum amount of weight that can be stacked on the aluminum plate is lbf.

A carbon steel ball with 27.00-mm diameter is pressed together with an aluminum ball with a 36.00-mm diameter by a force of 11.00 N. Determine the maximum shear stress and the depth at which it will occur for the aluminum ball. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pma 9 0.6 στ 24 0.4 Tmax 0.2 0 0.5a a 1.5a Z 2a 2.5a За Distance from contact surface The maximum shear stress is determined to be MPa. The depth in the aluminum ball at which the maximum shear stress will occur is determined to be [ mm.

Show all work please

Chapter 17 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

Ch. 17.1 - A round object of mass m and radius r is released...Ch. 17.1 - Prob. 17.2CQ Ch. 17.1 - Prob. 17.3CQ Ch. 17.1 - Prob. 17.4CQ Ch. 17.1 - Slender bar A is rigidly connected to a massless...Ch. 17.1 - A 200-kg flywheel is at rest when a constant 300...Ch. 17.1 - The rotor of an electric motor has an angular...Ch. 17.1 - Prob. 17.3P Ch. 17.1 - Two disks of the same material are attached to a...Ch. 17.1 - Prob. 17.5P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Find the centroidal radius of gyration for the rotor.

Solution Summary: The author calculates the centroidal radius of gyration for the rotor. The kinetic friction (M) produces a couple of magnitudes.

Concept explainers

Find the centroidal radius of gyration for the rotor.

Want to see the full answer?

Chapter 17 Solutions