Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Rotational Dynamics of a Bicycle wheel
Reply to thread
Message
[QUOTE="Special K213, post: 3583995, member: 367250"] [h2]Homework Statement [/h2] A bicycle wheel has a diameter of 63.8 cm and a mass of 1.79 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 117 N is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a 9.01-cm-diameter sprocket in order to give the wheel an acceleration of 4.53 rad/s2? (b) What force is required if you shift to a 5.60-cm-diameter sprocket? R=wheel radius R= .319m m= wheel mass m=1.79kg F=Resistive force tangent to tire F=117N r=radius of sprocket r=.0901m α=angular acceleration of the wheel α=4.53rad/s^2 τf= Torque from the bike wheel τext=external torque for sprocket [h2]Homework Equations[/h2] Torque τ=rF Moment of Interia of a hoop I=MR^2 Torque τ=Iα [h2]The Attempt at a Solution[/h2] First calculate torque on the bike wheel τf=.319m(117N)= 37.3Nm Due to laws of rotational dynamics and why the wheel is spinning relate the torque of the wheel to torque of sprocket to moment of interia of the wheel τext-τf=Iα Break down each term into quantities we already know τext=rF τf=37.3Nm I=MR^2 α=4.53rad/s^2 Thusly rF-37.3Nm=MR^2(4.53Rad/s^2) Rearrange and solve for F F=MR^2(α)+τf/r Input values F=(1.79kg)(.319)^2(4.53rad/s^2)+37.3Nm/.0901m F=423N Maybe I messed up somewhere but I've tried twice already submitting the answer online and this is wrong but that was my shot, thank you for helping. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Rotational Dynamics of a Bicycle wheel
Back
Top