Circular motion

Mass point moves moves uniformly in a circle with radius r = 3.4 m angular velocity ω = 3.6 rad/s. Calculate the period, frequency, and the centripetal acceleration of this movement.

Correct result:

T =  1.75 s
f =  0.57 Hz
a =  44.06 m/s2

Solution:

ω=2πf f=1/T T=2π3.6=1.75 s\omega = 2 \pi f \ \\ f = 1/T \ \\ T = \dfrac{ 2 \pi }{ 3.6} = 1.75 \ \text{s}
f=1/T=0.57 Hzf = 1/T = 0.57 \ \text{Hz}
a=rω2=3.43.62=44.06 m/s2a= r \omega^2 = 3.4 \cdot 3.6^2 = 44.06 \ \text{m/s}^2



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