Respuesta :
Part a)
as we know that angular speed is given as
[tex]\omega = 2 \pi f[/tex]
here we have
[tex]f_2 = 280 rpm = 4.67Hz[/tex]
[tex]f_1 = 130 rpm = 2.17 Hz[/tex]
now for angular acceleration we have
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
[tex]\alpha = \frac{2\pi(f_2 - f_1)}{\Delta t}[/tex]
[tex]\alpha = \frac{2\pi(4.67 - 2.17)}{4}[/tex]
[tex]\alpha = 3.93 rad/s^2[/tex]
Part b)
angular speed of the wheel after t = 2 s
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]\omega_f = 2\pi (2.17) + 3.93 (2)[/tex]
[tex]\omega_f = 21.5 rad/s[/tex]
now we have
tangential acceleration
[tex]a_t = r \alpha[/tex]
[tex]a_t = 0.70(3.93) = 2.75 m/s^2[/tex]
radial acceleration will be
[tex]a_r = \omega^2 r[/tex]
[tex]a_r = (21.5)^2(0.70) = 323.6 m/s^2[/tex]
A) Angular acceleration = 3.93 rad/s²
B) i) Tangential components = 2.75 m/s²
ii) Radial components = 323.6 m/s²
Given data :
Diameter of wheel = 70 cm = 0.7 m
Initial angular frequency ( f₁ ) = 130 rpm = 2.17 Hz
Final angular frequency ( f₂ ) = 280 rpm = 4.67 Hz
Time ( T ) = 4 secs
A) Calculate the angular acceleration
Angular acceleration ( ∝ ) = Δω / ΔT
= 15.71 / 4 = 3.93 rad/s²
where : Δω = 2π ( f₂ - f₁ ) = 2π ( 4.67 - 2.17 ) = 15.71
change in time ( ΔT ) = 4 secs
B) Determine the radial and tangential components of the linear acceleration after 2 secs
First step : calculate the angular speed after 2 secs
ωf = ωf₁ + ∝t
= 2π ( 2.17 ) + 3.93 * 2
= 21.5 rad/sec
Final step :
i) Tangential acceleration
at = r∝
at = 0.70 ( 3.93 ) = 2.75 m/s²
ii ) Radial acceleration
ar = ω²r
= ( 21.5 )² * ( 0.70 ) = 323.6 m/s²
Hence we can conclude that A) Angular acceleration = 3.93 rad/s²
B) i) Tangential components = 2.75 m/s², ii) Radial components = 323.6 m/s²
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