Question

A uniform rod of mass m=1.0 kg and length L=2 meters is free to rotate about its center as shown in the figure. A constant torque is applied by a constant force of magnitude F=5.0 N to one end of the rod, as shown. To be clear, the force is always perpendicular to the rod, which rotates about the axis indicated by the red line.a) If the rod is initially at rest, how long does it take to reach an angular velocity omega=30 radians/s. b) Assuming the torque stops when the angular velocity is 30 radians/s, what is the total rotational energy of the rod? c) The rod suddenly turns into a uniform sphere of radius R=1m and mass M=1.0 kg rotating about its center. Assuming angular momentum is conserved, what is the angular velocity of the sphere omegasphere?

Answer 1

a)

If the length is 2 meters and the rod rotates about its center, so 2 meters is the diameter, and the radius of rotation is 1 meter.

Then, if the force is 5 N, let's calculate the torque:

`\begin{gathered} \tau=F\cdot r \\ \tau=5\cdot1=5\text{ Nm} \end{gathered}`

Then, calculating the rotational inertia and the angular acceleration, we have:

`\begin{gathered} I=(1)/(2)mr^2 \\ I=(1)/(2)\cdot1\cdot1^2=0.5 \\ \\ \alpha=(\tau)/(I)=(5)/(0.5)=10 \end{gathered}`

The angular acceleration is 10 rad/s², so to reach an angular velocity of 30 rad/s, it takes 3 seconds.

b)

The rotational energy can be calculated with the formula below:

`\begin{gathered} E_k=(1)/(2)I\cdot\omega^2 \\ E_k=(1)/(2)\cdot0.5\cdot30^2 \\ E_k=225\text{ J} \end{gathered}`

c)

The angular momentum is given by:

`\begin{gathered} L=I\cdot\omega \\ L=0.5\cdot30 \\ L=15 \end{gathered}`

Then, since the rod turns into a sphere, the new rotational inertia is:

`\begin{gathered} I=(2)/(5)mr^2 \\ I=(2)/(5)\cdot1\cdot1^2 \\ I=(2)/(5)=0.4 \end{gathered}`

So the new angular velocity is:

`\begin{gathered} L=I\cdot\omega \\ 15=0.4\cdot\omega \\ \omega=(15)/(0.4)=37.5\text{ rad/s} \end{gathered}`