This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ghei (mg42382) HW 10 opyrchal (11130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A string is wound around a solid cylindrical spool of mass 8 . 3 kg and radius 0 . 18 m and is tied to the ceiling as shown in the figure. R M h=from bottom of cylinder to ground If the spool is released from rest and rolls on the string, and the distance to the floor is 8 . 75 m, how long will it take the spool to hit the floor? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 63663 s. Explanation: mg T summationdisplay F y = T M g = M ( a CM ) , (1) I = 1 2 M R 2 and = a CM /R The torque is summationdisplay center = + T R = I = parenleftbigg 1 2 M R 2 parenrightbigg a CM R (2) T = 1 2 M a cm By Newtons equation, summationdisplay F y = T M g = M ( a CM ) 1 2 M a CM M g = M a CM a CM = 2 3 g . This is constant acceleration, so h = 1 2 a CM t 2 t = radicalBigg 2 h a CM = radicalBigg 3 h g = 1 . 63663 s . alternative Solution: Let y be the height of the bottom of the spool at a given time. Using conservation of energy ( E f = E ), 1 2 I 2 + 1 2 M v 2 + M g y = M g h 1 2 parenleftbigg 1 2 M R 2 parenrightbigg parenleftBig v R parenrightBig 2 + 1 2 M v 2 = M g ( h y ) 3 4 v 2 = g ( h y ) v = radicalbigg 4 3 g ( h y ) . In the energy approach there is no guar- antee that the spool will move at a constant acceleration. The velocity is downward, so v = d y dt = radicalbigg 4 3 g ( h y ) 1 h y d y dt = radicalbigg 4 3 g . integraldisplay t h 1 h y d y dt dt = radicalbigg 4 3 g integraldisplay t h dt ghei (mg42382) HW 10 opyrchal (11130) 2 integraldisplay h ( h y )- 1 2 ( dy ) = radicalbigg 4 3 g integraldisplay t h dt 2 ( h y ) 1 2 vextendsingle vextendsingle vextendsingle h = radicalbigg 4 3 g t vextendsingle vextendsingle vextendsingle t h 2 h = radicalbigg 4 3 g t vextendsingle vextendsingle vextendsingle t h t h = radicalBigg 3 h g = radicalBigg 3 (8 . 75 m) 9 . 8 m / s 2 = 1 . 63663 s A quicker way to find t h Average velocity is defined by v = distance time = h t h Since the acceleration is constant, we have v = v + 0 2 = v 2 so t h = h v = 2 h v = 2 h radicalbigg 3 4 g = radicalBigg 3 h g keywords: 002 (part 1 of 2) 10.0 points A spool is unwound by a constant force pulling on a string. The radius of the spool is given in the figure. Assume: The moment of inertia of the spool about its center of mass is I = 4 5 m r 2 and the spool is rolling without slipping and starts from rest....
View Full Document