Can you offer some brief explanations along with the answers ? Thankyou aIAssign
ID: 1446130 • Letter: C
Question
Can you offer some brief explanations along with the answers ? Thankyou aIAssignmertp-ec The Expert TA 1 Human like x Class Management I Help Chapters 10 and 11 continued Begin Date: 3/20/2016 10:00:00 PM Due Date: 3/27/2016 11:59:00 PM End Date: 4/6/2016 11:59:00 PM (13%) Problem, 5: A turntable ofradius 25 cm and rotational inertia 0.0154kg m2 is spinning freely at 22.0 rpm about its central axis, with a 19.5g mouse on its outer edge. The mouse walks from the edge to the center. Assume the turntable-mouse system is isolated and put your answers in terms of the variables listed variable Description Value 0.25m Turntable inertia 19.5 Mouse mass nitial system rotationalinertia Final system rotational inertia Final system angular velocity Work done by mouse ©theexpert ta.com 17% Part (a) Treat the mouse as a point-particle, and define the turntable's initial rotation direction to be positive. Calculate the system moment of inertia before the mouse moves Ibefore in terms of the variables above. Grade Summary Potential 100% 7 8 9 Submissions Attempts remaining: 7 (3% per attempt) detailed view 001 Vr Submit Hint I give up Hints: 0% dediction per hit. Hints retnamang: 2 Feedback: 1%deduction per feedback. Submission History Answer Hints Feedback Totals Totals 0% | 0% 17% Part (b) Calculate the 17% Part (c) Assume the turntable-mouse system is isolated. I ind the system's final rotation sp 17% Part (d) Solve for the final rotational speed oy in rads. systems moment of inertia after the mouse moves to the center in terms of the variables above. system is isolated. Find the system's final rotation speed in terms of the variables above 17% Part (e) Ignore fiction within the turntable-mouse system. Determine the work done by the 17% Part (f) What is the numerical value of the work done in Joules? within the turntable-mouse system. Determine the work done by the mouse in terms of the variables above.
Explanation / Answer
(a)
initial moment of inertia of the system is
I_i = turntable rotational inertia + moment of inerita of mouse
=0.0154 kg m^2 + (19.5 * 10^-3 kg) (0.25 m)^2
=0.0166 kg m^2
(b)final moment of inertia
I_f = turntable rotational inertia + moment of inerita of mouse
=0.0154 kg m^2 + (19.5 * 10^-3 kg) (0 m)^2
=0.0154 kg m^2
(c)
from the conservation of angular momenum
I_i wi = I_f wf
wf = I_i wi /I_f
(d)wf = 22 rpm (0.0166 kg m^2/0.0154 kg m^2 =23.7 rpm
(e)
W = 1/2 I_f wf^2 - 1/2 I_i wi^2
(f)
W =1/2 (0.0154 kg m^2)( 23.7 ( 2 pi/ 60 s)^2-1/2 (0.0166 kg m^2)( 22 ( 2 pi/ 60 s)^2
=3.49 * 10^-3 J