Respuesta :
Answer:
[tex]a=368.97\ m/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega=0[/tex]
Acceleration of the wheel, [tex]\alpha =7\ rad/s^2[/tex]
Rotation, [tex]\theta=14\ rotation=14\times 2\pi =87.96\ rad[/tex]
Let t is the time. Using second equation of kinematics can be calculated using time.
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s[/tex]
Let [tex]\omega_f[/tex] is the final angular velocity and a is the radial component of acceleration.
[tex]\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s[/tex]
Radial component of acceleration,
[tex]a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2[/tex]
So, the required acceleration on the edge of the wheel is [tex]368.97\ m/s^2[/tex].
The radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s²
Using ω² = ω₀² + 2αθ, we find the final angular speed after 14 revolutions, ω where
- ω₀ = initial angular speed = 0 rad/s (since the wheel starts from rest),
- α = angular acceleration = 7 rad/s and
- θ = number of revolutions = 14 = 14 rev × 2π rad/rev = 87.965 rad.
So, substituting the values of the variables into the equation, we have
ω² = ω₀² + 2αθ,
ω² = (0 rad/s)² + 2 × 7 rad/s² × 87.965 rad.
ω² = 0 rad²/s² + 1231.504 rad²/s²
ω² = 1231.504 rad²/s²
ω = √(1231.504 rad²/s²)
ω = 35.09 rad/s
We know that the radial acceleration a = rω² where
- r = radius of wheel = diameter of wheel/2 = 60 cm/2 = 30 cm = 0.30 m and
- ω = final angular acceleration of wheel = 35.09 rad/s.
So, substituting the values of the variables into the equation, we have
a = rω²
a = 0.30 m × (35.09 rad/s)²
a = 0.30 m × 1231.504 rad²/s²
a = 369.45 m/s²
So, the radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s².
Learn more about radial acceleration here:
https://brainly.com/question/25243603
