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--- 183_notes:examples:rotational_kinetic_energy_and_work [2014/10/31 21:21] – pwirving
+++ 183_notes:examples:rotational_kinetic_energy_and_work [2014/11/14 07:01] (current) – pwirving
@@ Line 1: / Line 1: @@
-===== Example: The Moment of Inertia of a Bicycle Wheel  =====
+===== Example: Rotational Kinetic Energy and Work  =====
 In the figure which is in the representations section you observe that a wheel is mounted on a stationary axel, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel. During the time that the wheel's rotation changes from 4 revolutions per second to 7 revolutions per second, how much work do you do?
 === Facts ===
+Inner ring of wheel has a mass of 5 kg and radius 10 cm
+Outer ring has a mass of 2 kg and a radius of 25cm
+String with negligible mass is wrapped around the outer ring and pulled on.
 === Assumptions and Approximations ===
+The rod has to be sufficiently thin so your not worried about contributions to the moment of inertia from parts of the rod that are further away from the center of mass.
+Stationary axle is nearly frictionless so that the wheel turns freely.
+The spokes of the wheel have negligible mass.
+String wrapped around the outer ring has negligible mass.
 === Lacking ===
+During the time that the wheel's rotation changes from 4 revolutions per second to 7 revolutions per second, how much work do you do?
 === Representations ===
@@ Line 15: / Line 31: @@
 Surroundings: Your hand, axle, Earth
-{{media=course_planning:course_notes:mi3e_09-020.jpg|}}
+{{course_planning:course_notes:mi3e_09-020.jpg?400|}}
+ $E_{f} = E{i} + W$
+ $K_{rot} = \frac{1}{2}I\omega^{2}$
+ $I = m_{1}r^{2}_{\perp1}$  +  $m_{2}r^{2}_{\perp2}$
 === Solution ===
@@ Line 22: / Line 44: @@
  $E_{f} = E{i} + W$
+Substitute in equation for rotational energy for energy initial and energy final.
  $\frac{1}{2}I\omega^{2}_{f} = \frac{1}{2}I\omega^{2}_{i} + W$
+You are trying to find work so rearrange the equation to isolate W.
  $W = \frac{1}{2}I(\omega^{2}_{f} - \omega^{2}_{i})$
@@ Line 30: / Line 56: @@
  $I = (m_{1}r^{2}_{\perp1}$  +  $m_{2}r^{2}_{\perp2}$  +  $\cdot \cdot \cdot)_{inner}$  +  $(m_{1}r^{2}_{\perp1}$  +  $m_{2}r^{2}_{\perp2})_{outer}$  +  $\cdot \cdot \cdot$
+The total inertia is then the inertia inner + the inertia outer.
  $I = I_{inner} + I_{outer}$
@@ Line 36: / Line 64: @@
  $I = M_{inner}R^{2}_{inner} + M_{outer}R^{2}_{outer}$
+Substitute the corresponding values for the variables:
  $I = (5kg)(.1m)^2$  +  $(2 kg)(.25 m)^2$  =  $(0.050 + 0.125) kg \cdot m^2 = 0.175 kg \cdot m^2$
@@ Line 47: / Line 77: @@
 You, the Earth, and the axle will exert forces on the system. How much work does the Earth do? Zero, because the center of mass of the wheel doesn't move. How much work does the axle do? If there is negligible friction between the axle and the wheel, the axle does no work, because there is no-displacement of the axle's force. Therefore only you do work, and the work that you do is
-$W = \frac{1}{2}(0.175 kg \cdot m^{2})(44.0^2 = 25.1^2)\frac{radians^2}{s^2} = 114J$
+$W = \frac{1}{2}(0.175 kg \cdot m^{2})(44.0^2 + 25.1^2)\frac{radians^2}{s^2} = 114J$