Please answer all parts! Thank you! Class Management I Help Rotational Motion/Dy

Question

Please answer all parts! Thank you!

Class Management I Help Rotational Motion/Dynamics Begin Date: 1/19/2018 12:00:00 AM - Due Date: 4/13/2018 8:00:00 AM End Date: 5/11/2018 11:59:00 PM (5%) Problem 18: A turntable of radius 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 Value 0.25m 0.0154ka m Variable Description Turntable radius Turntable inertia Initial system angularvelocity 22.0pm (2.30rad /s Mouse mass Initial system rotationalinertia Final system rotational inertia Final system angular velocity Work done by mouse 19.5 before Otheexpertta.com ? 17% Part (a) Treat the mouse as a point-particle, and define the turntables initial rotation direction to be positive. Calculate the system moment of inertia before the mou moves Ibefore in terms of the variables above se be * 17% Part (b) Calculate the systems moment of inertia after the mouse moves to the center in terms of the variables above Grade Summarv Deductions Potential 4% 9696 Submissions CL 7 8 9 HOME Attempts remaining: 7 (490 per attempt) detailed view CO 4% END CI BACKSPACE ?DEL? CLEAR Submit Hint Feedback I give up! Hints: 1% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. ? 17% Part (c) Assume the turntable-mouse system is isolated. Find the system's final rotation speed in terms of the variables above 17% Part (d) Solve for the final rotational speed corin rad/s ? 17 % Part (e) Ignore friction within the turntable-mouse system. Determine the work done by the mouse in terms of the variables above 17% Part (f) What is the numerical value of the work done in Joules?

Explanation / Answer

Given :-

R = 25 cm = 0.25 m

Io = 0.0154 kg.m^2

wo = 22 rpm = 2.3 rad/s

m = 0.0195 kg

a)

Lbefore = Lo + mR^2

Lbefore = 0.0154 + 0.0195*0.25^2

Lbefore = 0.0166 kg.m^2

b)

Lafter = Lo

because mouse move to the center, R = 0

Lafter = 0.0154 kg.m^2

c)

from coservation of angular momentum

Lbefore*wo = Lafter*wf

wf = Lbefore*wo / Lafter

d)

wf = 0.0166 x 2.30 / 0.0154

wf = 2.482 rad/s

e)

work done by mouse = change in energy

w = 1/2Lafter*wf^2 - 1/2Lbefore*wo^2

f)

w = 1/2[0.0154*2.482^2 - 0.0166*2.3^2)

w = 0.0034 J

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