Question

A particle moves along a circle of radius (20/pi)m with constant tangential acceleration.if the velocity of the particle is 80m/s at the end of the second revolution after motion has begun,the tangential acceleration is

Answer 1

Solution :

Key Idea: The tangential acceleration in a circular path is the product of radius of circular path and angular acceleration. The tangential acceleration                 

aT=rα...(i)                

From 2nd equation of motion in rotational motion,                 
ω^2=ω^20+2αθ

                
Here,     

ω0=0,

ω=v/r=80/20/π=4πrad/s

θ=2×2πrad

So,

α=ω^2/2θ=(4π)^2/2×(2×2π)

=16π^2/8π=2π

                

Hence, from Eq. (i), we have                 
aT=rα=20/π×2π=40m/s^2

Answer 2

Answer:

40 m/s²

Explanation:

r = $\frac{20}{\pi }$ m

v = 80 m/s

θ = 2 revolutions = 4π rad.

Using Equation :

ω² = ω₀² + 2αθ     ..(ω₀=0)

ω² = 2αθ     ..(ω=v/r  &  a=rα)

a = v²/2rθ

a = 40 m/s²