R MasteringPhysics: Assignment 21 Google Chrome https://session.masteringphysics

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R MasteringPhysics: Assignment 21 Google Chrome https://session.masteringphysics.com/myct/itemview?assignmentProblemID 616643 Phys 2109-2110 Assignment 21 Item 5 Item 5 Part A A 111 g Frisbee is 18 cm in diameter and has half its mass spread uniformly in the disk What is the Frisbee's rotational inertia? and the other half concentrated in the rim. Express your answer using two significant figures. kg Subm My Answers Give Up Part B With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 670 rpm. What is the magnitude of the torque, assumed constant, that the student applies? Express your answer using two significant figures. Subm My Answers Give Up EE Search the web and Windows Signed in as Emily Wiedemann l Help l cose Resources previous l 5 of 7 l next Continue 11:40 PM 2016

Explanation / Answer

For rotation about the axes perpendicular to the disk and through the center, a uniform disk has moment of inertia 1/2 m R^2 . A ring has moment of inertia m r^2 .

Here the m = 1/2 M, with M = 0.111 kg .

I_frisbee = 1/2 m R^2 + m R^2
= 1/2 (1/2 M) R^2 + (1/2 M) R^2
= 3/4 M R^2
= 3/4 * 0.111 kg * (0.09 m)^2 = 6.74 *10^-4 kg m^2 .


For this uniform angular acceleration from rest, the angle of rotation a satisfies:
a(t) = 1/2 (dw/dt) t^2
and the angular velocity
w(t) = (dw/dt) * t

We know that when a = pi/2 radians, w = 2pi * 670/60 rad/s= 22.33pi rad/s

So we have the simultaneous equations

pi/2 = 1/2 (dw/dt) t^2
22.33 pi = (dw/dt) t

which gives the angular acceleration dw/dt as

dw/dt = 498.62 pi rad/s^2

Finally we then have for the torque
T = I dw/dt
= 6.74*10^-4 kg m^2 *498.62 pi /s^2 = 1.055 N m

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