The rigid object shown is rotated about an axis perpendicular to the paper and through point P. The total kinetic energy of the object as it rotates is equal to 1.4 J. If M = 1.3 kg and L = 0.50 m, what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles.
A)1.3 rad/s
B)1.5 rad/s
C)1.7 rad/s
D)1.2 rad/s
E)2.1 rad/s

Question

The rigid object shown is rotated about an axis perpendicular to the paper and through point P. The total kinetic energy of the object as it rotates is equal to 1.4 J. If M = 1.3 kg and L = 0.50 m, what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles. 
A)1.3 rad/s 
B)1.5 rad/s 
C)1.7 rad/s 
D)1.2 rad/s 
E)2.1 rad/s

Quizplus · Accepted Answer

The total kinetic energy of a rigid object rotating about an axis through point P is given by: 
$K = \frac{1}{2}I_P\omega^2$, 
where $I_P$ is the moment of inertia of the object about the axis through point P, and $\omega$ is the angular velocity of the object.

The moment of inertia of the rigid object about the given axis can be found by using the parallel-axis theorem: 
$I_P = I_{CM} + Md^2$, 
where $I_{CM}$ is the moment of inertia of the object about its own center of mass, $M$ is the total mass of the object, and $d$ is the perpendicular distance between the axis and the center of mass.

To find the moment of inertia about the given axis, we need to find the moments of inertia of the two masses (treated as particles) about their respective axes of rotation (through the center of mass of each particle), and then use the parallel-axis theorem to add them up.

For the mass at the end of the rod: 
$I_1 = \frac{1}{2}mr^2$, 
where $m$ is the mass of the particle and $r$ is the distance between the particle and the axis of rotation (which is equal to the length of the rod).

For the mass in the middle of the rod: 
$I_2 = \frac{1}{12}ml^2 + \frac{1}{4}m(\frac{r}{\sqrt{2}})^2 = \frac{1}{12}ml^2 + \frac{1}{8}mr^2$, 
where $l$ is the length of the rod and $\frac{r}{\sqrt{2}}$ is the distance between the particle and the axis of rotation. We use the parallel-axis theorem to add the contribution of the mass to the moment of inertia about the axis through point P. 
$I_2 = \frac{1}{12}ml^2 + \frac{1}{8}mr^2 + md^2$, 
where $d$ is the distance between the center of mass of the rod (which is at the middle of the rod) and the axis through point P. We can use geometry to find that $d = \frac{r}{\sqrt{2}} - \frac{L}{2}$.

Thus, the total moment of inertia about the axis through point P is: 
$I_P = I_1 + I_2 = \frac{1}{2}mr^2 + \frac{1}{12}ml^2 + \frac{5}{8}mr^2 + M(\frac{r}{\sqrt{2}} - \frac{L}{2})^2$
$= (\frac{7}{12}mr^2 + \frac{1}{12}ml^2 + \frac{1}{2}M(\frac{r}{\sqrt{2}})^2) + M(\frac{L}{2\sqrt{2}})^2$ 
$= (\frac{7}{12} + \frac{1}{6})mr^2 + (\frac{1}{12} + \frac{1}{2})ml^2$ 
$= \frac{3}{2}mr^2 + \frac{1}{4}ml^2$.

Plugging in the given values, we get: 
$K = \frac{1}{2}I_P\omega^2 = \frac{1}{2}(\frac{3}{2}mr^2 + \frac{1}{4}ml^2)\omega^2 = 1.4 J$.

Solving for $\omega$, we get: 
$\omega = \sqrt{\frac{2K}{(\frac{3}{2}mr^2 + \frac{1}{4}ml^2)}} = \sqrt{\frac{2(1.4)}{(\frac{3}{2})(1.3)(0.25^2) + (\frac{1}{4})(1.3)(0.5^2)}}$
$\approx 1.7 rad/s$.

Therefore, the answer is C.

The Rigid Object Shown Is Rotated About an Axis Perpendicular