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?
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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
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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
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Answer:
40 m/s²
Explanation:
r = 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
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