Answer with Explanation:
We are given that
Radius of disk=r=7.98 cm=0.0798 m
1 m=100 cm
a.Angular speed=1250rev/min
[tex]2\pi[/tex] rad=1 rev
[tex]1250 rev/min=1250\times 2\pi[/tex] rad/min
1 min=60 sec
Angular speed=[tex]\omega=\frac{1250\times 2\pi}{60} rad/s[/tex]
Angular speed=[tex]\omega=130.9 rad/sec[/tex]
b.r=3.1 cm=[tex]\frac{3.1}{100}=0.031m[/tex]
1 m=100 cm
[tex]v=r\omega=0.031\times 130.9=4.058m/s[/tex]
Hence, the tangential speed=4.058m/s
c.Radial acceleration=[tex]a=r\omega^2=(130.9)^2(0.0798)=1367.36m/s^2[/tex]
Hence, the radial acceleration of a point on the rim=[tex]1367.36 m/s^2[/tex]
d.Time, t=1.9 s
Total distance=[tex]d=r\omega t[/tex]
[tex]d=0.0798\times 130.9\times 1.9[/tex]
[tex]d=19.8 m[/tex]
