 #1
ubiquinone
 43
 0
Hi there, I have a question regarding energy that I'm not sure if I have solved correctly. I was wondering if anyone here may please have a look. Thanks!
Question: A 2 kg mass is on the end of a massless rigid rod which pivots freely about one end. The mass is moving in a vertical circle with a speed of 6 m/s at P when it reaches Q, its speed is 8.00 m/s. Find a) the length of the rod b) force exerted by the rod on the mass when the mass is at P.
Diagram:
The total energy at P: [tex]E_T=\frac{1}{2}(2kg)(6m/s)^2+(2kg)(9.8N/kg)(l\sin 30l\sin 310)[/tex]
The total energy at A:
[tex]E_T=\frac{1}{2}(2kg)(8m/s)^2[/tex]
The total energy at P and A are equal, so solving for [tex]l[/tex], [tex]\displaystyle l=\left(\frac{6436}{19.6(\sin 30\sin 310)}\right)m=2.58m[/tex]
b) At P: [tex]E_T=F\cdot l\Rightarrow F=\frac{E_T}{l}[/tex]
Therefore, [tex]\displaystyle F=\left(\frac{64J}{2.58m}\right)=24.8N[/tex]
Question: A 2 kg mass is on the end of a massless rigid rod which pivots freely about one end. The mass is moving in a vertical circle with a speed of 6 m/s at P when it reaches Q, its speed is 8.00 m/s. Find a) the length of the rod b) force exerted by the rod on the mass when the mass is at P.
Diagram:
Code:
P
\
\
\
30 deg \
\
__________ +
*

 *

 *

 *
40 deg
 * Q
The total energy at A:
[tex]E_T=\frac{1}{2}(2kg)(8m/s)^2[/tex]
The total energy at P and A are equal, so solving for [tex]l[/tex], [tex]\displaystyle l=\left(\frac{6436}{19.6(\sin 30\sin 310)}\right)m=2.58m[/tex]
b) At P: [tex]E_T=F\cdot l\Rightarrow F=\frac{E_T}{l}[/tex]
Therefore, [tex]\displaystyle F=\left(\frac{64J}{2.58m}\right)=24.8N[/tex]