Respuesta :
9) 1.55 rad/s^2
The angular acceleration of the disk is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 28 rad/s[/tex] is the final angular speed
[tex]\omega_i=0[/tex] is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
Substituting into the equation, we find:
[tex]\alpha = \frac{28.1 rad/s - 0}{18.1 s}=1.55 rad/s^2[/tex]
10) 253.9 rad
The angular displacement of the disk during this time interval is given by the equation:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where
[tex]\omega_i=0[/tex] is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
[tex]\alpha=1.55 rad/s^2[/tex] is the angular acceleration
Substituting into the equation, we find:
[tex]\theta = 0 + \frac{1}{2}(1.55 rad/s^2)(18.1 s)^2=253.9 rad[/tex]
11) [tex]0.428 kg m^2[/tex]
The moment of inertia of a disk rotating about its axis is given by
[tex]I=\frac{1}{2}mR^2[/tex]
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
[tex]I=\frac{1}{2}(9.5 kg)(0.3 m)^2=0.428 kg m^2[/tex]
12) 167.8 J
The rotational energy of the disk is given by
[tex]E_R = \frac{1}{2}I\omega^2[/tex]
where
[tex]I=0.428 kg m^2[/tex] is the moment of inertia
[tex]\omega[/tex] is the angular speed
At the beginning, [tex]\omega_i = 0[/tex], so the rotational energy is
[tex]E_i = \frac{1}{2}(0.428 kg m^2)(0)^2 = 0[/tex]
While at the end, the angular speed is [tex]\omega=28 rad/s[/tex], so the rotational energy is
[tex]E_f = \frac{1}{2}(0.428 kg m^2)(28 rad/s)^2=167.8 J[/tex]
So, the change in rotational energy of the disk is
[tex]\Delta E= E_f - E_i = 167.8 J - 0 = 167.8 J[/tex]
13) [tex]0.47 m/s^2[/tex]
The tangential acceleration can be found by using
[tex]a_t = \alpha r[/tex]
where
[tex]\alpha = 1.55 rad/s^2[/tex] 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
[tex]a_t = (1.55 rad/s^2)(0.3 m)=0.47 m/s^2[/tex]
14) [tex]58.8 m/s^2[/tex]
The radial (centripetal acceleration) is given by
[tex]a_r = \omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed, which is half of its final value, so
[tex]\omega=\frac{28 rad/s}{2}=14 rad/s[/tex]
r is the distance of the point from the centre (as before, r = R = 0.3 m)
Substituting numbers into the equation,
[tex]a_r = (14 rad/s)^2 (0.3 m)=58.8 m/s^2[/tex]
15) 4.2 m/s
The tangential speed is given by:
[tex]v=\omega r[/tex]
where
[tex]\omega = 28 rad/s[/tex] 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,
[tex]r=\frac{R}{2}=\frac{0.3 m}{2}=0.15 m[/tex]
So the tangential speed is
[tex]v=(28 rad/s)(0.15 m)=4.2 m/s[/tex]
16) 77.0 m
The total distance travelled by a point on the rim of the disk is
[tex]d=ut + \frac{1}{2}a_t t^2[/tex]
where
u = 0 is the initial tangential speed
t = 18.1 s is the time
[tex]a_t = 0.47 m/s^2[/tex] is the tangential acceleration
Substituting into the equation, we find
[tex]d=0+\frac{1}{2}(0.47 m/s^2)(18.1 s)^2=77.0 m[/tex]
