A yoyo is constructed with a clever clutch mechanism in its axle that allows it to rotate freely and "sleep" when its angular speed is greater than a certain critical value. When the yo-yo's angular speed falls below this value the clutch engages, causing the yo-yo to climb the string to the user's hand. If the moment of inertia of the yo-yo is 7.4×10-5 kg•m2 its mass is 0.11 kg, and the string is 0.98 m long, what is the smallest angular speed that will allow the yo-yo to return to the user's hand? (in rad/s)
I didn't really know how to approach this question. Using the moment of inertia and mass, I thought I could find the radius to be 0.02594 m...but I didn't know where to go from there or if that will actually help me. How can I find theta and/or the time to solve this problem? Thanks in advance for any and all help!
2007-03-19 09:35:47 · 2 answers · asked by eureka4sureka 1 in Science & Mathematics ➔ Physics
Since there is no friction, you can use conservation of energy. In this case it says that kinetic energy at the speed of clutch engagement = potential energy when it rises to the user's hand.
KE = (1/2)*I*w^2 = m*g*h
2007-03-19 10:51:17 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
each and each 2d, the shaft advantageous properties 3.00 rad/s in extra angular speed, so the angular speed after 18 seconds is: 3.00 rad/s^2 * 18 s = fifty 4 rad/s. to locate the finished perspective of rotation after 18 seconds, you combine the speed with appreciate to time, from t=0 to t=18 seconds: critical((3 rad/s)*t, t, 0, 18) = (3 rad/s)*t^2 / 2, evaluated from 0 to 18 = 486 rad
2016-10-02 09:50:56 · answer #2 · answered by zaragosa 4 · 0⤊ 0⤋