A disk with mass m = 10.5 kg and radius R = 0.35 m begins at rest and accelerate

Question

A disk with mass m = 10.5 kg and radius R = 0.35 m begins at rest and accelerates uniformly for t = 17.4 s, to a final angular speed of = 29 rad/s.

1)

What is the angular acceleration of the disk?

2)

What is the angular displacement over the 17.4 s?

3)

What is the moment of inertia of the disk?

4)

What is the change in rotational energy of the disk?

5)

What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

6)

What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

7)

What is the final speed of a point on the disk half-way between the center of the disk and the rim?

8)

What is the total distance a point on the rim of the disk travels during the 17.4 seconds?

Explanation / Answer

Given that

A disk with mass (m) = 10.5 kg and radius(R ) = 0.35 m

Initial speed of the disk is (vi) =0

The time ( t) = 17.4 s

The final angular speed of (f) = 29 rad/s

1)

The angular acceleation is given by

alpha =wf-wi/t =29-0/17.4s =1.667rad/s2

2)

Displacement is given by

Deltatheta =wi+(1/2)alpha*t2 =0+0.5*1.667rad/s2*(17.4)2 =252.350rad

3)

The moment of inertia of the disk is

I =(1/2)mR2 =0.5*10.5*(0.35)2 =0.643kg.m2

4)

Change in kinetic energy is

DeltaK =KEf-KEi =(1/2)Iwf2 -(1/2)Iwi2 =(1/2)Iwf2 =0.5*0.643*(29)2 =270.434J

5)

The tangential component of the acceleration of a point on the rim of the disk(at) =alpha*R =1.667*0.35 =0.583m/s2

6)

The radial acceleration is given by ar =w2*R =(29/2)2(0.35) =73.587m/s2

7)

Now vt =w*R =(14.5)(0.35) =5.075m/s

8)

The total distance at a point is given by

DeltaS =Deltatheta*R =(252.350)(0.35) =88.322m

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