Question

A strap is passing over wheel of radius 30 cm during the time the wheel moving with the initial constant velocity of 2 rev/ seconds comes to rest the strap covers a distance of 25 m the deacceleration of the wheel in rad per second is

Answer 1

r = radius of the wheel = 30 cm = 0.30 m

C = circumference of the wheel = distance traveled by wheel in one revolution = 2πr

D = total distance traveled by strap = 25 m

N = total number of revolutions by wheel

total number of revolutions by wheel is given as

N = D/C = 25/(2πr)

θ = angular displacement = 25/(2πr)  = (25/(2πr)) ( 2π)  rad = 25/r = 25/0.30 = 83.33 rad

w₀ = initial angular velocity of wheel = 2 rev/s = 2 (2π) rad/s  = 12.56 rad/s

w = final angular velocity after wheel stops = 0 rad/s

α = angular acceleration

Using the equation

w² = w²₀  + 2 αθ

0² = 12.56² + 2 α (83.33)

α = - 0.95 rad/s²

Answer 2

r = radius of the wheel = 30 cm = 0.30 m

C = circumference of the wheel = distance traveled by wheel in one revolution = 2πr

D = total distance traveled by strap = 25 m

N = total number of revolutions by wheel

total number of revolutions by wheel is given as

N = D/C = 25/(2πr)

θ = angular displacement = 25/(2πr)  = (25/(2πr)) ( 2π)  rad = 25/r = 25/0.30 = 83.33 rad

w₀ = initial angular velocity of wheel = 2 rev/s = 2 (2π) rad/s  = 12.56 rad/s

w = final angular velocity after wheel stops = 0 rad/s

α = angular acceleration

Using the equation

w² = w²₀  + 2 αθ

0² = 12.56² + 2 α (83.33)

α = - 0.95 rad/s²

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