A uniform rod (mass = 1.5 kg) is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a horizontal position. What is the angular speed of the rod when the rod makes an angle of 30° with the horizontal? (The rotational inertia of the rod about the pin is 2.0 kg⋅m²). A)2.2 rad/s B)3.6 rad/s C)2.7 rad/s D)3.1 rad/s E)1.8 rad/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. When the rod is released from rest, all energy is initially in the form of gravitational potential energy:

$U_i = mgh = (1.5\ 	ext{kg})(9.81\ 	ext{m/s}^2)(2.0\ 	ext{m}) = 29.4\ 	ext{J}$

At an angle of 30°, the rod has lost some height and gained some rotational kinetic energy:

$U_f = \frac{1}{2}I\omega^2 + mgh_f$

where $h_f=2.0	ext{m}-2.0	ext{m}\cos(30^\circ)=1.732	ext{m}$ is the height of the rod at 30°. Substituting in the values given and solving for $\omega$ gives:

$\omega = \sqrt{\frac{2(U_f-U_i)}{I}} = \sqrt{\frac{2[(1/2)(2.0	ext{ kg}\cdot	ext{m}^2)(\omega^2) + (1.5\ 	ext{kg})(9.81\ 	ext{m/s}^2)(1.732\ 	ext{m}) - 29.4\ 	ext{J}]}{2.0 	ext{ kg}\cdot	ext{m}^2}}$

Simplifying and solving for $\omega$ gives:

$\omega = \sqrt{\frac{147.22}{2.0}} = 6.07\ 	ext{rad/s}$

However, we need to find the angular velocity when the angle is 30°, not the angular velocity at the bottom. We can use conservation of angular momentum to relate these two values:

$L_i = L_f$

$(0)(I) = (mgh)(L/2)$

$L = 2mgh = 2(1.5\ 	ext{kg})(9.81\ 	ext{m/s}^2)(2.0\ 	ext{m})\sin(30^\circ) = 58.0\ 	ext{J}\cdot	ext{s}$

$L=I\omega$

$\omega = \frac{L}{I} = \frac{58.0\ 	ext{J}\cdot	ext{s}}{2.0\ 	ext{kg}\cdot	ext{m}^2} = 29.0\ 	ext{rad/s}$

Finally, we can find the angular velocity at 30° using conservation of energy again:

$U_f = \frac{1}{2}I\omega^2 + mgh_f$

$\omega = \sqrt{\frac{2(U_f-U_i)}{I}} = \sqrt{\frac{2[(1/2)(2.0	ext{ kg}\cdot	ext{m}^2)(29.0\ 	ext{rad/s})^2 + (1.5\ 	ext{kg})(9.81\ 	ext{m/s}^2)(1.732\ 	ext{m}) - 29.4\ 	ext{J}]}{2.0 	ext{ kg}\cdot	ext{m}^2}} = 2.7\ 	ext{rad/s}$

So the answer is choice C, 2.7 rad/s.

A Uniform Rod (Mass = 1