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A WHEEL ROTATES WITH CONSTANT ANGULAR ACCELERATION OF 2 rad/SEC²
HAS INITIAL ANGULAR VELOCITY AS 4 rad/sec² THEN NO OF REVOLUTION IT MAKE IN FIRST 10 SECONDS WILL BE??
Answers
Is q m sbse pehle hum angular acceleration nikalenge wo 4 aayega equation use krk usk baad isme hum angular velocity nikalenge equation se w=w•+alpha×t se iska answer 40Π aajayega
α = angular acceleration = rad/sec²
t = time = 5 sec.
ω₀ = initial angular velocity
= 0 rad/sec as the body is at rest at t = 0 sec.
θ = total angle rotated from t=0 to t,
= 50 * 2π radians.
Total angle rotated = θ = ω₀ t + 1/2 * α * t²
(It's like: total distance = s = u t + 1/2 * a * t²)
So 100 π = 0 * 5 + 1/2 * α * 5²
α = 8π rad/sec²
= 8π /(2π) revolutions/sec² , as 1 revolution = 2π rad.
= 4 revolutions/sec².
α = 4 rev/sec.
-- direct calculation in rev/sec²
Total angle rotated = θ = ω₀ t + 1/2 * α * t²
50 rev = 0 * 5 + 1/2 * α * 5²
α = 4 rev/sec²
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