|
|
| *Vultures Knob>>>Bike Lights |
Pulling a String Adds Energy to a Wheel? |
A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia I=kmr^2, where m is its mass, r is its radius, and k is a dimensionless constant between zero and one. The wheel is rotating counterclockwise with angular velocity w0, when at time t=0 someone starts pulling the string with a force of magnitude f. Assume that the string does not slip on the wheel. A)Suppose that after a certain time tL, the string has been pulled through a distance L. What is the final rotational speed wFinal of the wheel? B) What is the instantaneous power P delivered to the wheel via the force F at time t=0? i got F*r*w0 for b but wasnt correct im stuck A) Torque = I alpha. Moment of inertia (I) is given. Torque is force times radius. So you can find angular acceleration (alpha) angular displacement (theta)= distance pulled / r. Use (omega final)^2 - (omega inital)^2= 2 (alpha)(delta theta) to get your rotational speed (omega). B) Your answer is correct. I guess |
| Tags |
| Bike Tours Trail Map Bike Pedals Bike Lock Bike Lights Bike Helmet Quad Bike Pocket Bike Mini Bike |
Bike Categories--Copyright/IP Policy--Contact Webmaster |