Respuesta :
Answer:
a) [tex]-22.5 rad/s^2[/tex]
b) 30.6 revolutions
c) 4.13 s
d) 52.9 m
e) 25.6 m/s
Explanation:
a)
The relationship between linear acceleration and angular acceleration for an object in circular motion is given by
[tex]a=\alpha r[/tex]
where
[tex]a[/tex] is the linear acceleration
[tex]\alpha[/tex] is the angular acceleration
r is the radius of the motion of the object
For the tires of the car in this problem, we have:
[tex]a=-6.2 m/s^2[/tex] is the linear acceleration (the car is slowing down, so it is a deceleration, therefore the negative sign)
r = 0.275 m is the radius of the tires
Solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{a}{r}=\frac{-6.2}{0.275}=-22.5 rad/s^2[/tex]
b)
To solve this part of the problem, we can use the suvat equation for the rotational motion, in particular:
[tex]\omega^2 - \omega_0^2 = 2\alpha \theta[/tex]
where:
[tex]\omega[/tex] is the final angular velocity
[tex]\omega_0[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
[tex]\theta[/tex] is the angular displacement
Here we have:
[tex]\omega=0[/tex] (the tires come to a stop)
[tex]\omega_0 = 93 rad/s[/tex]
[tex]\alpha = -22.5 rad/s^2[/tex]
Solving for [tex]\theta[/tex], we find the angular displacement:
[tex]\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{0^2-(93)^2}{2(-22.5)}=192.2 rad[/tex]
And since 1 revolution = [tex]2\pi rad[/tex],
[tex]\theta=\frac{192.2}{2\pi}=30.6 rev[/tex]
c)
To solve this part, we can use another suvat equation:
[tex]\omega=\omega_0 + \alpha t[/tex]
where in this case, we have:
[tex]\omega=0[/tex] is the final angular velocity, since the tires come to a stop
[tex]\omega_0 = 93 rad/s[/tex] is the initial angular velocity
[tex]\alpha=-22.5 rad/s^2[/tex] is the angular acceleration
t is the time
Solving for t, we can find the time required for the tires (and the car) to sopt:
[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{0-93}{-22.5}=4.13 s[/tex]
d)
The car travels with a uniformly accelerated motion, so we can find the distance it covers by using the suvat equations for linear motion:
[tex]s=vt-\frac{1}{2}at^2[/tex]
where:
v = 0 is the final velocity of the car (zero since it comes to a stop)
t = 4.13 s is the time taken for the car to stop
[tex]a=-6.2 m/s^2[/tex] is the deceleration for the car
s is the distance covered during this motion
Therefore, substituting all values and calculating s, we find the distance covered:
[tex]s=0-\frac{1}{2}(-6.2)(4.13)^2=52.9 m[/tex]
e)
The relationship between angular velocity and linear velocity for a rotational motion is given by
[tex]v=\omega r[/tex]
where
v is the linear velocity
[tex]\omega[/tex] is the angular speed
r is the radius of the circular motion
In this problem:
[tex]\omega_0 = 93 rad/s[/tex] is the initial angular speed of the tires
r = 0.275 m is the radius of the tires
Therefore, the initial velocity of the car is:
[tex]u=\omega_0 r = (93)(0.275)=25.6 m/s[/tex] is the initial velocity of the car
