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3 years ago

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Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at

rest. The disks are brought together by applying a horizontal force of magnitude 20 N to the axle of disk A. Assume that μk = 0.15 between the disks. Neglect bearing friction.
Required:
Determine:
a. the angular acceleration of each disk.
b. the final angular velocity of each disk.

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

Answer:

a. the angular acceleration of each disk;

For disk A, the angular acceleration is 9.36 rad/s²

For disk B, the angular acceleration is 28.57 rad/s²

b. the final angular velocity of each disk;

For disk A, the final angular velocity is 250.57 rpm

For disk A, the final angular velocity is 334.00 rpm

Explanation:

While slipping occurs, and upward friction force F is applied to disk A, and downward friction force F to disk B.

Assumption;

Radius of disk A is 80 mm

Radius of disk B is 60 mm.

Pls find the explanation of the question in the attached files as typing it here may be difficult due to the equations and diagrams involved.

garri49 [273]3 years ago

4 0

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

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