Purchase Solution

Dynamics Problem - Torque, Acceleration, and Angular Velocity

Not what you're looking for?

Ask Custom Question

An electric motor is running at 1200rpm when its power is cut off for 40 seconds, then turned back on. If the rotor weighs 60lb, has a radius of gyration of 8in, and experiences a power-on net torque of 100lb/in and power-off friction torque of 8lb/in, the rotational speed in rpm of the motor 2 seconds after the power is turned back on is closest to which answer?

a. 624
b. 1085
c. 21.5
d. 114

*Please show your work.

Attachments
Purchase this Solution
Solution Summary

This solution is provided in 168 words. It uses step by step calculations for the power off and power on phases for the system to find out angular acceleration and angular velocity.

Solution Preview

initial rotation speed wo = 1200 rpm= 1200/60 = 20 rps = 20*2*pi= 40pi rad/s
power cut duration t = 40 sec

mass of the rotor m = 60 lb
radius of gyration k = 8 in = 8/12 = 2/3 ft
power on net ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
Recent Feedback
  • " In question 2, you incorrectly add in the $3.00 dividend that was just paid to determine the value of the stock price using the dividend discount model. In question 4 response, it should have also been recognized that dividend discount models are not useful if any of the parameters used in the model are inaccurate. "
  • "feedback: fail to recognize the operating cash flow will not begin until the end of year 3."
  • "Answer was correct"
  • "Great thanks"
  • "Perfect solution..thank you"
Purchase this Solution
Free BrainMass Quizzes
Variables in Science Experiments

How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.

The Moon

Test your knowledge of moon phases and movement.

Classical Mechanics

This quiz is designed to test and improve your knowledge on Classical Mechanics.

Basic Physics

This quiz will test your knowledge about basic Physics.

Introduction to Nanotechnology/Nanomaterials

This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.

View More Free Quizzes
Related BrainMass Solutions

  • Moment of inertia: Force of fish on the line; how much line unwinds?

    (1)] To find the amount of line that unwinds use the fact that the line is collected at the reel so its length will follow velocity and acceleration given by v = r x omega [radius times angular velocity] and a = r x alpha Therefore if the length

  • Rotational Dynamics and Net Torque

    It uses revolutions to calculate radius, and then calculates angular acceleration and torque.

  • torque and angular momentum

    What angular momentum will it have 10s after a 2.5 N.m torque is applied to it? , so the angular acceleration is So its angular velocity after 10 seconds is The angular momentum is 3.

  • Rotational Motion: Torque, angular acceleration, velocity

    A block is attached to the string wrapped on a cylindrical pulley, the torque, angular velocity and angular acceleration is calculated.

  • Pulley System - Torque, Inertia, Acceleration, Velocity, and Kinetic Energy

    It uses simple calculations to find net torque, angular acceleration, velocity, and kinetic energy.

  • Dynamics Problem - Torque and Angular Acceleration

    35549 Dynamics Problem - Torque and Angular Acceleration A slender rod of mass Mr has a disk of mass Mo fastened to it at its end. It is pinned at 0 - a "frictionless" pin.

  • Rotational Dynamics

    acceleration (α) Substituting values: - 0.3 = 0.027α α = - 0.3/0.027 = - 11.11 rad/s2 Initial frequency of rotation of the wheel = f = 900 rpm or 900/60 = 15 rps Initial angular velocity ω0 = 2Пf = 2x3.14x15 = 94.2 rad/sec Final angular

  • A flywheel coming to rest.

    Final angular velocity= initial angular velocity + acceleration * time Here the torque is opposing the rotation and the acceleration will be replaced by deceleration Final angular velocity =( initial angular velocity) + (deceleration * time) 0 radians

  • Centripetal Acceleration Rotor

    (a) Current angular speed = (b) Angular acceleration = Change in angular velocity/ time interval = (2094.4 - 0)/(5*60) = 6.98 rad/s2 (c) I = mr2 = 2*0.12 = 0.02 kg-m2 Torque = I*Angular acceleration = 0.02*6.98 = 0.1396 N-m.

View More Related Solutions