9) 1.55 rad/s^2
The angular acceleration of the disk is given by
![\alpha = \frac{\omega_f - \omega_i}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%20-%20%5Comega_i%7D%7Bt%7D)
where
is the final angular speed
is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
Substituting into the equation, we find:
![\alpha = \frac{28.1 rad/s - 0}{18.1 s}=1.55 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B28.1%20rad%2Fs%20-%200%7D%7B18.1%20s%7D%3D1.55%20rad%2Fs%5E2)
10) 253.9 rad
The angular displacement of the disk during this time interval is given by the equation:
![\theta = \omega_i t + \frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
where
is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
is the angular acceleration
Substituting into the equation, we find:
![\theta = 0 + \frac{1}{2}(1.55 rad/s^2)(18.1 s)^2=253.9 rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%281.55%20rad%2Fs%5E2%29%2818.1%20s%29%5E2%3D253.9%20rad)
11) ![0.428 kg m^2](https://tex.z-dn.net/?f=0.428%20kg%20m%5E2)
The moment of inertia of a disk rotating about its axis is given by
![I=\frac{1}{2}mR^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7DmR%5E2)
where in this case we have
m = 9.5 kg is the mass of the disk
R = 0.3 m is the radius of the disk
Substituting numbers into the equation, we find
![I=\frac{1}{2}(9.5 kg)(0.3 m)^2=0.428 kg m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7D%289.5%20kg%29%280.3%20m%29%5E2%3D0.428%20kg%20m%5E2)
12) 167.8 J
The rotational energy of the disk is given by
![E_R = \frac{1}{2}I\omega^2](https://tex.z-dn.net/?f=E_R%20%3D%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2)
where
is the moment of inertia
is the angular speed
At the beginning,
, so the rotational energy is
![E_i = \frac{1}{2}(0.428 kg m^2)(0)^2 = 0](https://tex.z-dn.net/?f=E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.428%20kg%20m%5E2%29%280%29%5E2%20%3D%200)
While at the end, the angular speed is
, so the rotational energy is
![E_f = \frac{1}{2}(0.428 kg m^2)(28 rad/s)^2=167.8 J](https://tex.z-dn.net/?f=E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%280.428%20kg%20m%5E2%29%2828%20rad%2Fs%29%5E2%3D167.8%20J)
So, the change in rotational energy of the disk is
![\Delta E= E_f - E_i = 167.8 J - 0 = 167.8 J](https://tex.z-dn.net/?f=%5CDelta%20E%3D%20E_f%20-%20E_i%20%3D%20167.8%20J%20-%200%20%3D%20167.8%20J)
13) ![0.47 m/s^2](https://tex.z-dn.net/?f=0.47%20m%2Fs%5E2)
The tangential acceleration can be found by using
![a_t = \alpha r](https://tex.z-dn.net/?f=a_t%20%3D%20%5Calpha%20r)
where
is the angular acceleration
r is the distance of the point from the centre of the disk; since the point is on the rim,
r = R = 0.3 m
So the tangential acceleration is
![a_t = (1.55 rad/s^2)(0.3 m)=0.47 m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%20%281.55%20rad%2Fs%5E2%29%280.3%20m%29%3D0.47%20m%2Fs%5E2)
14) ![58.8 m/s^2](https://tex.z-dn.net/?f=58.8%20m%2Fs%5E2)
The radial (centripetal acceleration) is given by
![a_r = \omega^2 r](https://tex.z-dn.net/?f=a_r%20%3D%20%5Comega%5E2%20r)
where
is the angular speed, which is half of its final value, so
![\omega=\frac{28 rad/s}{2}=14 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B28%20rad%2Fs%7D%7B2%7D%3D14%20rad%2Fs)
r is the distance of the point from the centre (as before, r = R = 0.3 m)
Substituting numbers into the equation,
![a_r = (14 rad/s)^2 (0.3 m)=58.8 m/s^2](https://tex.z-dn.net/?f=a_r%20%3D%20%2814%20rad%2Fs%29%5E2%20%280.3%20m%29%3D58.8%20m%2Fs%5E2)
15) 4.2 m/s
The tangential speed is given by:
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where
is the angular speed
r is the distance of the point from the centre of the disk, so since the point is half-way between the centre of the disk and the rim,
![r=\frac{R}{2}=\frac{0.3 m}{2}=0.15 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7BR%7D%7B2%7D%3D%5Cfrac%7B0.3%20m%7D%7B2%7D%3D0.15%20m)
So the tangential speed is
![v=(28 rad/s)(0.15 m)=4.2 m/s](https://tex.z-dn.net/?f=v%3D%2828%20rad%2Fs%29%280.15%20m%29%3D4.2%20m%2Fs)
16) 77.0 m
The total distance travelled by a point on the rim of the disk is
![d=ut + \frac{1}{2}a_t t^2](https://tex.z-dn.net/?f=d%3Dut%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_t%20t%5E2)
where
u = 0 is the initial tangential speed
t = 18.1 s is the time
is the tangential acceleration
Substituting into the equation, we find
![d=0+\frac{1}{2}(0.47 m/s^2)(18.1 s)^2=77.0 m](https://tex.z-dn.net/?f=d%3D0%2B%5Cfrac%7B1%7D%7B2%7D%280.47%20m%2Fs%5E2%29%2818.1%20s%29%5E2%3D77.0%20m)