Class Management Help April 4 Begin Date: 4/1/2016 12:00:00 AM Due Date: 4/4/201
ID: 1453611 • Letter: C
Question
Class Management Help April 4 Begin Date: 4/1/2016 12:00:00 AM Due Date: 4/4/2016 2:00:00 AM End Date: 4/4/2016 2:00:00 AM (20%) Problem 4: An ice skater is spinning at 6.4 rev/s and has a moment of inertia of 0.24 kg m Randomized Variables 6.4 rev/s 0.75 rev/s 2.5 rev/s I 0.24 kg m 13 A 33% Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.4 rev/s Grade Summary Deductions 0% Potential 100% tano (1 7 8 9 HOME sino cos Submissions A Attempts remaining 5 cotano asin acos0 4 5 6 10% per attempt) L sinh0 atan() acotan 1 2 3 detailed view cosh() tanho cotanho o Degrees Radians Feedback give up! Submit Hint Hints 0% deduction per hint. Hints remaining: Feedback 0% deduction per feedback. A 33% Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared if his rate of rotation decreases to 0.75 rev/s. A 33% Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 2.5 rev/s. What is the magnitude of the average torque that was exerted, in N m, if this takes 13 s? tent 2016 Expert TA, LLCExplanation / Answer
(a)angular momentum=Iw
if w=6.4rev/s=40.21 radian/sec
angular momentum=0.24*40.21=9.65 kg m^2/s
(b) if w is decreased to 0.75 rev /sec
by conservation of angular momentum
I1*w1=I2*w2
I2=I1*w1/w2=2.048 kgm^2
(c) average torque=I*angular acceleration
angular acceleration=2*pi(2.5-6.4)/13=1.885 radian/sec^2
average torque=0.24*1.885=0.45 Nm