Question

A disk with mass m = 9.5 kg and radius R = 0.3 m begins at rest and accelerates uniformly for t = 18.1 s, to a final angular speed of ω = 28 rad/s. 9) What is the angular acceleration of the disk? rad/s2 10) What is the angular displacement over the 18.1 s? rad 11) What is the moment of inertia of the disk? kg-m2 12) What is the change in rotational energy of the disk? J 13) What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 14) What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 15) What is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s 16) What is the total distance a point on the rim of the disk travels during the 18.1 seconds? m

Answer 1

9) 1.55 rad/s^2

The angular acceleration of the disk is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 28 rad/s$ is the final angular speed

$\omega_i=0$ is the initial angular speed (the disk starts from rest)

t = 18.1 s is the time interval

Substituting into the equation, we find:

$\alpha = \frac{28.1 rad/s - 0}{18.1 s}=1.55 rad/s^2$

10) 253.9 rad

The angular displacement of the disk during this time interval is given by the equation:

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i=0$ is the initial angular speed (the disk starts from rest)

t = 18.1 s is the time interval

$\alpha=1.55 rad/s^2$ is the angular acceleration

Substituting into the equation, we find:

$\theta = 0 + \frac{1}{2}(1.55 rad/s^2)(18.1 s)^2=253.9 rad$

11) $0.428 kg m^2$

The moment of inertia of a disk rotating about its axis is given by

$I=\frac{1}{2}mR^2$

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

$I=\frac{1}{2}(9.5 kg)(0.3 m)^2=0.428 kg m^2$

12) 167.8 J

The rotational energy of the disk is given by

$E_R = \frac{1}{2}I\omega^2$

where

$I=0.428 kg m^2$ is the moment of inertia

$\omega$ is the angular speed

At the beginning, $\omega_i = 0$, so the rotational energy is

$E_i = \frac{1}{2}(0.428 kg m^2)(0)^2 = 0$

While at the end, the angular speed is $\omega=28 rad/s$, so the rotational energy is

$E_f = \frac{1}{2}(0.428 kg m^2)(28 rad/s)^2=167.8 J$

So, the change in rotational energy of the disk is

$\Delta E= E_f - E_i = 167.8 J - 0 = 167.8 J$

13) $0.47 m/s^2$

The tangential acceleration can be found by using

$a_t = \alpha r$

where

$\alpha = 1.55 rad/s^2$ 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

$a_t = (1.55 rad/s^2)(0.3 m)=0.47 m/s^2$

14) $58.8 m/s^2$

The radial (centripetal acceleration) is given by

$a_r = \omega^2 r$

where

$\omega$ is the angular speed, which is half of its final value, so

$\omega=\frac{28 rad/s}{2}=14 rad/s$

r is the distance of the point from the centre (as before, r = R = 0.3 m)

Substituting numbers into the equation,

$a_r = (14 rad/s)^2 (0.3 m)=58.8 m/s^2$

15) 4.2 m/s

The tangential speed is given by:

$v=\omega r$

where

$\omega = 28 rad/s$ 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,

$r=\frac{R}{2}=\frac{0.3 m}{2}=0.15 m$

So the tangential speed is

$v=(28 rad/s)(0.15 m)=4.2 m/s$

16) 77.0 m

The total distance travelled by a point on the rim of the disk is

$d=ut + \frac{1}{2}a_t t^2$

where

u = 0 is the initial tangential speed

t = 18.1 s is the time

$a_t = 0.47 m/s^2$ is the tangential acceleration

Substituting into the equation, we find

$d=0+\frac{1}{2}(0.47 m/s^2)(18.1 s)^2=77.0 m$