Answer:
a) ![-22.5 rad/s^2](https://tex.z-dn.net/?f=-22.5%20rad%2Fs%5E2)
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
![a=\alpha r](https://tex.z-dn.net/?f=a%3D%5Calpha%20r)
where
is the linear acceleration
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:
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
, we find the angular acceleration:
![\alpha = \frac{a}{r}=\frac{-6.2}{0.275}=-22.5 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7Ba%7D%7Br%7D%3D%5Cfrac%7B-6.2%7D%7B0.275%7D%3D-22.5%20rad%2Fs%5E2)
b)
To solve this part of the problem, we can use the suvat equation for the rotational motion, in particular:
![\omega^2 - \omega_0^2 = 2\alpha \theta](https://tex.z-dn.net/?f=%5Comega%5E2%20-%20%5Comega_0%5E2%20%3D%202%5Calpha%20%5Ctheta)
where:
is the final angular velocity
is the initial angular velocity
is the angular acceleration
is the angular displacement
Here we have:
(the tires come to a stop)
![\omega_0 = 93 rad/s](https://tex.z-dn.net/?f=%5Comega_0%20%3D%2093%20rad%2Fs)
![\alpha = -22.5 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20-22.5%20rad%2Fs%5E2)
Solving for
, we find the angular displacement:
![\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{0^2-(93)^2}{2(-22.5)}=192.2 rad](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B%5Comega%5E2-%5Comega_0%5E2%7D%7B2%5Calpha%7D%3D%5Cfrac%7B0%5E2-%2893%29%5E2%7D%7B2%28-22.5%29%7D%3D192.2%20rad)
And since 1 revolution =
,
![\theta=\frac{192.2}{2\pi}=30.6 rev](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B192.2%7D%7B2%5Cpi%7D%3D30.6%20rev)
c)
To solve this part, we can use another suvat equation:
![\omega=\omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega%3D%5Comega_0%20%2B%20%5Calpha%20t)
where in this case, we have:
is the final angular velocity, since the tires come to a stop
is the initial angular velocity
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:
![t=\frac{\omega-\omega_0}{\alpha}=\frac{0-93}{-22.5}=4.13 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Comega-%5Comega_0%7D%7B%5Calpha%7D%3D%5Cfrac%7B0-93%7D%7B-22.5%7D%3D4.13%20s)
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:
![s=vt-\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dvt-%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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
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:
![s=0-\frac{1}{2}(-6.2)(4.13)^2=52.9 m](https://tex.z-dn.net/?f=s%3D0-%5Cfrac%7B1%7D%7B2%7D%28-6.2%29%284.13%29%5E2%3D52.9%20m)
e)
The relationship between angular velocity and linear velocity for a rotational motion is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where
v is the linear velocity
is the angular speed
r is the radius of the circular motion
In this problem:
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:
is the initial velocity of the car