Question

A DC motor is specified to have a stall torque, Tstall of 10 in-lbf and a no-load speed, ωNL = 100 RPM. a. What is the maximum power output of this motor in in-lbf/s? In Horsepower? In Watts? b. What is the equation of torque, T, in terms of angular speed, ω?

          A DC motor is specified to have a stall torque, Tstall of 10 in-lbf and a no-load speed, ωNL = 100 RPM. 

a. What is the maximum power output of this motor in in-lbf/s? In Horsepower? In Watts? 

b. What is the equation of torque, T, in terms of angular speed, ω?
        
Show more…

Added by Joshua O.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
A DC motor is specified to have a stall torque, Tstall of 10 in-lbf and a no-load speed, ωNL = 100 RPM. a. What is the maximum power output of this motor in in-lbf/s? In Horsepower? In Watts? b. What is the equation of torque, T, in terms of angular speed, ω?
Close icon
Play audio
Feedback
Powered by NumerAI
Kathleen Carty David Collins
Ivan Kochetkov verified

Paul Gabriel and 96 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
a-motor-has-a-back-emf-of-110-mathrmv-and-an-armature-current-of-90-a-when-running-at-1500-rpm-deter

A motor has a back emf of $110 \mathrm{~V}$ and an armature current of 90 A when running at 1500 rpm. Determine the power and the torque developed within the armature. Power $=($ Armature current $)($ Back emf $)=(90 \mathrm{~A})(110 \mathrm{~V})=9.9 \mathrm{~kW}$ From Chapter 10, power $=\tau \omega$ where $\omega=2 \pi f=2 \pi(1500 \times 1 / 60)$ $\mathrm{rad} / \mathrm{s}$ $$ \text { Torque }=\frac{\text { Power }}{\text { Angular speed }}=\frac{9900 \mathrm{~W}}{(2 \pi \times 25) \mathrm{rad} / \mathrm{s}}=63 \mathrm{~N} \cdot \mathrm{m} $$

Schaum’s Outline of College Physics

a-motor-runs-at-20-mathrmrev-mathrms-and-supplies-a-torque-of-75-mathrmn-cdot-mathrmm-what-horsepowe

A motor runs at $20 \mathrm{rev} / \mathrm{s}$ and supplies a torque of $75 \mathrm{~N} \cdot \mathrm{m}$. What horsepower is it delivering? Using $\omega=20 \mathrm{rev} / \mathrm{s}=40 \pi \mathrm{rad} / \mathrm{s}$, we have $$ \mathrm{P}=\tau \omega=(75 \mathrm{~N} \cdot \mathrm{m})(40 \pi \mathrm{rad} / \mathrm{s})=9.4 \mathrm{~kW}=13 \mathrm{hp} $$

Schaum’s Outline of College Physics

a-permanent-magnet-dc-motor-has-an-armature-resistance-of-05-and-when-a-voltage-of-120-v-is-applied-to-the-motor-it-reaches-a-steadystate-speed-of-rotation-of-20revs-and-draws-40a-what-will-07127

A permanent magnet DC motor has an armature resistance of 0.5 Ω. When a voltage of 120 V is applied to the motor, it reaches a steady-state speed of rotation of 20 rev/s and draws 40 A. What will be: (a) the power input to the motor? (b) the power loss in the armature? (c) the torque generated at that speed?

Sri K.


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,241 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,732 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,290 solutions

*

Transcript

-
00:01 Hi, first question you have a motor has a back amf of 110 volts and an amateur current 90 ampers and it's running at 1 ,500 revolutions minutes wanted to determine the power and the talk so for the power is equal to the back amf times an amateur current this is equal to 110 times 90 which is 9 ,900 watts, to 9 .9 kilowatts.
01:07 Now for the torque, we have that the power is equal to times the angular velocity.
01:21 And we're giving the angular velocity as 1 ,500 revolutions per minute.
01:30 So we're going to convert this to radiance per second...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever