Question

A 4.0 cm diameter crankshaft turns at 2400 revolutions per minute.What is the speed of a point on the surface of the crankshaft?

Answer 1

(4*400/3600)*2*pi

Answer 2

solve it and u get the answer in m/sec

Answer 3

d = 2r
d = 4.0 cm = 4.0x10^-2 m
r = d/2 = (4.0x10^-2 m)/2 = 2.0x10^-2m

rpm=1/60 rev/s
2400 rpm =2400/60 rev/s = 40 rev/s

\[1 rev = 2\pi rad\]\[40 rev/s = 80 \pi rad/s\]

\[s=r \theta\]\[ds/dt=r (d \theta/dt)\]\[v=r \omega\]\[v = (2.0\times10^{-2} m) \times (80\pi rev/s) = 5.0 m/s\]