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
