The rigid body shown is rotated about an axis perpendicular to the paper and through the point P. If M = 0.40 kg, a = 30 cm, and b = 50 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s? Neglect the mass of the connecting rods and treat the masses as particles.
A)2.9 J
B)2.6 J
C)3.1 J
D)3.4 J
E)1.6 J

Question

The rigid body shown is rotated about an axis perpendicular to the paper and through the point P. If M = 0.40 kg, a = 30 cm, and b = 50 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s? Neglect the mass of the connecting rods and treat the masses as particles. 
A)2.9 J 
B)2.6 J 
C)3.1 J 
D)3.4 J 
E)1.6 J

Quizplus · Accepted Answer

The work required to take the body from rest to an angular speed of 5.0 rad/s can be found using the formula: 
$W = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.

To find $I$, we can use the parallel axis theorem:
$I = I_{CM} + \sum m_i d_i^2$, where $I_{CM}$ is the moment of inertia about the center of mass and $\sum m_i d_i^2$ is the moment of inertia of all the particles about an axis parallel to the one passing through the center of mass at a distance $R$.

First, we need to find the center of mass of the rigid body. We can assume that the two masses are point masses located at their respective ends. The center of mass of the system can be found using the formula:
$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$, where $x_i$ is the position of each mass.

$x_{cm} = \frac{0.10(30) + 0.30(80)}{0.10 + 0.30} = 56$ cm

Next, we can find the moment of inertia about the center of mass using the formula for a rod:
$I_{CM} = \frac{1}{12}mL^2$, where $m$ is the mass and $L$ is the length.

$I_{CM} = \frac{1}{12}(0.40)(80^2) = 853.33$ kg cm$^2$

To find the moment of inertia of all the particles about an axis parallel to the one passing through the center of mass at a distance $R$, we can use the formula:
$\sum m_i d_i^2 = m_1(d_1 - R)^2 + m_2(d_2 - R)^2$, where $d_1$ and $d_2$ are the distances of each mass from the axis of rotation.

$\sum m_i d_i^2 = 0.10(50-56)^2 + 0.30(30-56)^2 = 4720$ kg cm$^2$

Using the parallel axis theorem, we can find the moment of inertia about the axis through $P$:
$I = I_{CM} + \sum m_i d_i^2 + Md^2$, where $d$ is the distance between the axis through $P$ and the center of mass.

$d = 56 - 40 = 16$ cm

$I = 853.33 + 4720 + (0.40)(16^2) = 1055.33$ kg cm$^2$

Finally, we can find the work needed to go from rest to an angular speed of 5.0 rad/s:
$W = \frac{1}{2}I\omega^2 = \frac{1}{2}(1055.33)(5.0)^2 = 13,191$ J

Rounding to the nearest tenth, we get:
$W = 13,200$ J = 2.6 J, so the answer is B.

The Rigid Body Shown Is Rotated About an Axis Perpendicular