(8%) Problem 9: A turntable of radius 25 cm and rotational inertia 0.0154kg-m2 i
ID: 778367 • Letter: #
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(8%) Problem 9: A turntable of radius 25 cm and rotational inertia 0.0154kg-m2 is spinning freely at 22.0 rom 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 0.0154kg m Variable Description Turntable radius Turntable inertia Initial system angular velocity Mouse mass Initial system rotational inertia Final system rotational inertia Final system angular velocity Work done by mouse 22.0rpm (2.30rad/s 19.5 before Otheexpertta.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 Deductions Potential beforeI 0% 100% Submissions Attempts remaining: 10 (0% per attempt) detailed view Vo Submit Hint Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. 17% Part (b) Calculate the systems moment of inertia after the mouse moves to the center in terms of the variables above 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 a), in 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
a) I_before = Io + m*R^2
b) I_after = Io
c) Apply conservation of angular mmentum
If*I_after = I_before*wo
If = I_before*wo/I_after
= ( Io + m*R^2 )*wo/Io
d) wf = (0.0154 + 19.5*10^-3*0.25^2)*2.30/0.0154
= 2.48 rad/s
e) Workdone by mouse = increase in kinetic energy
= (1/2)*I_after*wf^2 - (1/2)*I_before*wo^2
f) Workdone by mouse = (1/2)*0.0154*2.48^2 - (1/2)*(0.0154 + 19.5*10^-3*0.25^2)*2.30^2
= 0.00340 J
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