A disk with mass m = 10.1 kg and radius R = 0.33 m begins at rest and accelerate

Question

A disk with mass m = 10.1 kg and radius R = 0.33 m begins at rest and accelerates uniformly for t = 16.4 s, to a final angular speed of = 34 rad/s.

1)

What is the angular acceleration of the disk?  

rad/s2

2)

What is the angular displacement over the 16.4 s?

rad

3)

What is the moment of inertia of the disk?

kg-m2

4)

What is the change in rotational energy of the disk?

J

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?  

m/s2

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?  

m/s2

7)

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

m/s

8)

What is the total distance a point on the rim of the disk travels during the 16.4 seconds?

Explanation / Answer

m (mass)= 10.1 kg , R = radius = 0.33m

t ( time ) = 16.4 sec, w0 = initial anular velocity = 0rad/s, w = final angular velocity = 34 rad/s

i) angular acceleration, @ = dw / dt

= 34 / 16.4 = 2.07 rad/s2

2) angular displacement , theta = w0*t + ( 1/2)*@*t2

= 0 + ( 1/2)*@*t2

= ( 1/2)*2.07*16.42 = 278.3 rad

3)moment of interia, I = (1/2)*m*R2

= (1/2)*10.1*0.332

= 0.549 kg - m2

4)change in rotational energy, Ek = ( 1/2)*I*w2 - ( 1/2)*I*w02

= ( 1/2)*0.549*(34)2 - 0

= 317.3 J

5) tangential acceleration , a = @*R

= 2.07*0.33 = 0.68 m/s2

6)radial acceleration , a = w2*R

= (34)2*0.33 = 381.4 m/s2

7) final speed, V = w*R

= 34*0.33 = 11.22 m/s

8) total distance S = theta*R

= 278.3*0.33 = 91.83 m

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