Question

A thin rod of length 1.3 m and mass 250 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 5.88 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises. (a) Number Enter your answer for part (a) in accordance to the question statement Units Choose the answer for part (a) from the menu in accordance to the question statement

Answer 1

(a) 2.42 J

The kinetic energy of a rotating object is given by:

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

where

I is the moment of inertia

$\omega$ is the angular speed

Here we have

$\omega=5.88 rad/s$ at the lowest point of the trajectory

While the moment of inertia of a rod rotating around one end is

$I=\frac{1}{3}ML^2 = \frac{1}{3}(0.250 kg)(1.3 m)^2=0.14 kg m^2$

And substituting in the previous formula, we find the kinetic energy at the lowest position:

$K=\frac{1}{2}(0.14 kg m^2)(5.88 rad/s)^2=2.42 J$

(b) 0.99 m

According to the law of conservation of energy, the total mechanical energy (sum of kinetic energy and potential energy) must be conserved:

$E=K+U$

At the lowest point, we can take the potential energy as zero, so the mechanical energy is just kinetic energy:

$E=K=2.42 J$

At the highest point in the trajectory, the rod is stationary, so the kinetic energy will be zero, and the mechanical energy will simply be equal to the gravitational potential energy:

$E=2.42 J = U = mgh$

where h is the heigth of the centre of mass of the rod with respect to the lowest point of the trajectory. Solving for h, we find

$h=\frac{E}{mg}=\frac{2.42 J}{(0.250 kg)(9.81 m/s^2)}=0.99 m$