Answer:
a) α₂ = 0.6 rad /s² , b) t = 13.1 s
Explanation:
a) This is an angular kinematics exercise. Let's analyze the situation when the two wheels are in contact and without sliding between them the linear speed is the same. Let's write your expressions
Rubber wheel v₁ = w₁ r₁
Ceramic wheel v₂ = w₂ r₂
v₁ = v₂
w₁ r₁ = w₂ r₂
w₂ = w₁ r₁ / r₂
We use the angular kinematic equation for the rubber wheel
w₁ = w₀ + alf₁ t
w₁ = α₁ t
we replace
w₂ = r₁ / r₂ (α₁ t)
w₂ = 0.020 / 0.25 7.5 t
w₂ = 0.6 t
therefore,
α₂ = 0.6 rad /s²
b) let's reduce to the SI system
w₂ = 75 rpm (2π rad / 1 rev) (1min / 60s) = 7.854 rad / s
We clear the equation and calculate
t = w₂ / alf₂
t = 7.854 / 0.6
t = 13.1 s
The angular acceleration of the pottery wheel is 0.6 rad /s² .
The time it takes the pottery wheel to reach its required speed of 75 rpm is 13.1 s.
This can be defined as the the rate of change of angular velocity with time.
Both wheels have the following kinematics.
Rubber wheel v₁ = w₁ r₁
Ceramic wheel v₂ = w₂ r₂
v₁ = v₂
w₁ r₁ = w₂ r₂
w₂ = w₁ r₁ / r₂
Using angular kinematic equation:
w₁ = w₀ + alf₁ t
w₁ = α₁ t
Substitute values into the equation
w₂ = r₁ / r₂ (α₁ t)
w₂ = 0.020 / 0.25 7.5 t
w₂ = 0.6 t
Hence, α₂ = 0.6 rad /s²
To calculate time, we convert to standard unit
w₂ = 75 rpm (2π rad / 1 rev) (1min / 60s) = 7.854 rad / s
Substitute values into the equation
t = w₂ / alf₂
t = 7.854 / 0.6
t = 13.1 s
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