Agyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.
(a) what is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?
(b) what is the radial acceleration of this point when the flywheel is spinning at full speed?
(c) through what distance does a point on the rim move during the spin-up?

Part a)

Part b)

Part c)

Explanation:

Part a)

as we know that angular acceleration of the wheel is given as

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

Part b)

frequency of the wheel at maximum speed is given as

now we know that

now radial acceleration is given as

Part c)

total angular displacement of the point on rim is given as

here we know that

now angular displacement will be

now the distance moved by the point on the rim is given as

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm