Respuesta :
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
A: The constant angular acceleration of the wheel is 44.34 rad/s2.
B: The constant angular velocity of the wheel is 195.13 rad/s.
C: The constant angular velocity of the wheel at t=0 s is 345.91 rad/s.
D: The angle at wheel turned between t = 0 and t = 3.4 s is 919.7272 rad.
E: The time at 6.65 seconds, the radial acceleration of the wheel will be equal to the acceleration due to gravity.
What is angular velocity?
The angular velocity can be defined as the time rate at which an object rotates or revolves about its axis.
Given that the radius r of the wheel is 0.230 m and tangential speed is 51.0 m/s and tangential acceleration is 10.2 m/s2.
Part A: Angular Acceleration
The constant angular acceleration of the wheel is calculated as given below.
[tex]\alpha = \dfrac {a_t}{r}[/tex]
[tex]\alpha = \dfrac {10.2}{0.230}[/tex]
[tex]\alpha = 44.34 \;\rm rad/s^2[/tex]
The constant angular acceleration of the wheel is 44.34 rad/s2.
Part B: Angular Velocity
The tangential speed v is calculated as given below.
[tex]v = v_t + a_t(t-t_0)[/tex]
[tex]v = 51.0 + (-10.2)(3.40 - 2.80)[/tex]
[tex]v = 44.88 \;\rm m/s[/tex]
So the angular velocity is given as,
[tex]\omega = \dfrac {v}{r}[/tex]
[tex]\omega = \dfrac {44.88}{0.230}[/tex]
[tex]\omega = 195.13 \;\rm rad/s[/tex]
The constant angular velocity of the wheel is 195.13 rad/s.
Part C: Angular Velocity at t=0
The tangential speed v is calculated as given below.
[tex]v = v_t + a_t(t-t_0)[/tex]
[tex]v = 51.0 + (-10.2)(0 - 2.80)[/tex]
[tex]v = 79.56 \;\rm m/s[/tex]
So the angular velocity is given as,
[tex]\omega = \dfrac {v}{r}[/tex]
[tex]\omega = \dfrac {79.56}{0.230}[/tex]
[tex]\omega = 345.91\;\rm rad/s[/tex]
The constant angular velocity of the wheel at t=0 s is 345.91 rad/s.
Part D:
The angle at wheel turned is calculated by the formula given below.
[tex]\theta = \omega \Delta t + \dfrac {1}{2} \alpha \Delta t^2[/tex]
[tex]\Delta t = 3.4 -0 = 3.4 \;\rm s[/tex]
[tex]\theta =195.13\times 3.4 + \dfrac {1}{2} \times \times 44.34\times 3.4^2[/tex]
[tex]\theta = 663.442 + 256.2852[/tex]
[tex]\theta = 919.7272 \;\rm rad[/tex]
The angle at wheel turned between t = 0 and t = 3.4 s is 919.7272 rad.
Part E:
The radial acceleration is equal to the acceleration due to gravity. This can be written as,
[tex]\alpha _r = \omega_0^2 r[/tex]
[tex]g = \omega_0^2 r[/tex]
[tex]\omega_0 = \sqrt{\dfrac {g}{r}}[/tex]
[tex]\omega_0 = \sqrt{\dfrac {9.8}{0.230}}[/tex]
[tex]\omega_0 = 6.52 \;\rm rad /s[/tex]
The time is taken by the wheel to reach from [tex]\omega[/tex] to [tex]\omega _0[/tex] is calculated as given below.
[tex]\omega _0 = \omega + \alpha t[/tex]
[tex]t = \dfrac {\omega_0 -\omega }{\alpha }[/tex]
[tex]t = \dfrac { 6.52-345.91}{-51}[/tex]
[tex]t = 6.65 \;\rm s[/tex]
The time at 6.65 seconds, the radial acceleration of the wheel will be equal to the acceleration due to gravity.
To know more about the angular velocity, follow the link given below.
https://brainly.com/question/1980605.
