- #1
GoldShadow
- 23
- 0
Homework Statement
Problem reads:
The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kgm^2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.
Homework Equations
[tex]E=\frac{1}{2}I\omega^{2}+m_{1}gh_{1}+m_{2}gh_{2}[/tex]
[tex]E_{initial}=E_{final}[/tex]
[tex]v=\omega r[/tex]
The Attempt at a Solution
I said [tex]m_{1}[/tex] was the 4.00 kg block and [tex]m_{2}[/tex] was the 2.00 kg block. Setting the initial and final energies of the system equal, I got:
[tex]m_{1}gh_{10}+m_{2}gh_{20}+\frac{1}{2}I\omega^{2}_{0}=m_{1}gh_{1}+m_{2}gh_{2}+\frac{1}{2}I\omega^{2}[/tex]
Since initial angular momentum is zero and so is the height of block 2, and in the final state, height of block 2 is zero, this simplifies to:
[tex]m_{1}gh_{10}=m_{2}gh_{2}+\frac{1}{2}I\omega^{2}[/tex]
Plugging in numbers:
[tex](4.00)(9.8)(5.00)=(2.00)(9.8)(5.00)+\frac{1}{2}(0.480)\omega^{2}[/tex]
I got [tex]\omega=20.21 rad/s[/tex]. Then using [tex]v=\omega r[/tex] I just plugged in the radius and the angular velocity I just found to get v=3.23 m/s
This is not the right answer according to the homework website unfortunately... help would be appreciated!
Last edited: