Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
The strength of the force of friction depends on two factors; the type of surfaces involves and the force that the surfaces are being push together
