A 111 g Frisbee is 20 cm in diameter and has half its mass spread uniformly in t

Question

A 111 g Frisbee is 20 cm in diameter and has half its mass spread uniformly in the disk and the other half concentrated in the rim. Part A What is the Frisbee's rotational inertia? Express your answer using two significant figures. I = 8.3×104 kgm2 SubmitMy AnswersGive Up Correct Part B With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 600 rpm . What is the magnitude of the torque, assumed constant, that the student applies? Express your answer using two significant figures.

Explanation / Answer

Diameter of disk = 0.20 m

Radius = 0.10 m

The problem can be divided in two parts first a disk of uniformly spread mass of 55.5 g

and second a ring of 55.5 g and radius 0.10m

Moment of inertia of disk =

Here M =55.5 g = 0.0555 kg

I1= 1/2(0.0555 *0.10*0.10)

= 0.0002775kg m2

Moment of inertia of ring of mass 55.5 g and radius 0.10 m

=Mr2

= 0.000555 kg m2

The total moment of inertia is:

I = 8.3×104 kgm2

Final velocity = 600 rpm = 600*2 radian per minute

=( 600*2)/60 radian per second

=62.8 radian per second

Now work done = change in kinetic energy

Kinetic energy was initially 0 whic becomes 1/2 I2

So, = 1/2 I2

= 1/2 I2/

Here = 90 degree = /4 radian

Putting the values we get

= 2.1 N.m

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