Question

If the length of spoke of a wirelis 30 om from artis of rotation
Then find the acceleration of a point on tip of the spoke,
Wheel is rotating a 120 rpm (Neglect the circumference of
rubber tube and tyres constraints)

Answer 1

Answer:

Here, no of spokes, N=10

length of the spoke, l=radius,r=0.5m,

Frequency, n=120 rpm =2rps.

B=0.4 gauss =0.4×10−4 tesla, e = ?

As the wheel rotates, linear velocity of spoke end at the rim=rω and linear velocity of spoke end at the axle =0

∴ Average linear velocity, υ=0+rω2=12rω

As e.m.f. induced acorss the ends of each spoke

e=Bυl=B.(12rω).l

=B12r2πnl=Brπnl

e=0.4×10−4×0.5×227×2×0.5

=6.28×10−5 volt

As all spokes are connected in parallel between the axle and the rim, therefore net e.m.f. induced is the same as that induced acorss the ends of each spoke.