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 rotational kinetic energy of a rigid object is given by:
$K_{rot} = \frac{1}{2} I \omega^2$
where $I$ is the moment of inertia and $\omega$ is the angular velocity.

To find the moment of inertia of the object about the axis of rotation at point P, we can use the parallel axis theorem:
$I_{P} = I_{COM} + Md^2$
where $I_{COM}$ is the moment of inertia about the center of mass, $M$ is the total mass, and $d$ is the distance between the center of mass and the axis of rotation.

From inspection, the center of mass of the object is at the midpoint of the rod, so we can treat the object as two point masses, one at the center of mass and one at the point P. The mass at the center of mass is $M_{COM} = 1.0$ kg and the mass at P is $M_P = 0.3$ kg.

The distance between the center of mass and the axis of rotation is $L/2 = 0.25$ m. The distance between point P and the center of mass is also $L/2$.

The moment of inertia about the center of mass of each mass is:
$I_{COM} = \frac{1}{2}m r^2$
where $m$ is the mass and $r$ is the distance from the mass to the center of mass. For the center of mass mass, $r = L/4 = 0.125$ m, so $I_{COM} = (1.0/2)(1.0)(0.125)^2 = 0.0078$ kg m$^2$. For the point P mass, $r = L/4 = 0.125$ m, so $I_{COM} = (1.0/2)(0.3)(0.125)^2 = 0.0029$ kg m$^2$.

Using the parallel axis theorem, the moment of inertia about the axis of rotation at point P is:
$I_P = I_{COM} + Md^2 = 0.0078 + 0.3(0.25)^2 + 0.0029 = 0.012$

Substituting into the expression for rotational kinetic energy and solving for $\omega$:
$K_{rot} = \frac{1}{2} I \omega^2$
$\omega = \sqrt{\frac{2K_{rot}}{I}} = \sqrt{\frac{2(1.4)}{0.012}} = 1.7$ rad/s

Therefore, the answer is choice C.

The Rigid Object Shown Is Rotated About an Axis Perpendicular