Question

a+particle+moves+along+a+circle+of+radius+(20/π)+with+constant+tangential+acceleration+.+If+the+velocity+of+the+particle+is+80+m/s+at+the+end+of+the+second+revolution+after+motion+has+began.+What+is+the+tangential+acceleration?

Answer 1

A particle is moving along a circle of radius 20/\pi20/π with constant tangential acceleration. The velocity of the particle is 80m/s at the end of 2nd  revolution after motion has began. Then the tangential acceleration is

ANSWER

r=\dfrac{20}{\pi}r=  

π

20

​  

 

In two revolution , distance is s=2\times 2\pi r=80ms=2×2πr=80m

Now, v^2=2asv  

2

=2as for zero initial speed.

\implies a=\dfrac{v^2}{2s}=\dfrac{80\times 80}{2\times 80}⟹a=  

2s

v  

2

 

​  

=  

2×80

80×80

​  

 

\implies a=40m/s^2⟹a=40m/s  

2

Explanation:hi

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