- #1
Antepolleo
- 40
- 0
I can't for the life of me figure out what I'm doing wrong here. Here's the problem:
A 5.40 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 7.00 cm and mass 2.00 kg.
Ok... the first question is:
What is the net torque on the system about the point O?
Here's my take on the situation... the only torque being applied to the system is from the tension T of the string.
T = tension of string
m = mass of counterweight
a = acceleration of counterweight
R = radius of wheel in meters
M = mass of pulley
[tex]T - mg = -ma[/tex]
[tex]T = m(g - a)[/tex]
[tex]
\begin{equation*}
\begin{split}
\sum\tau &= TR \\
&= mR(g - a)
\end{split}
\end{equation*}
[/tex]
Also...
[tex]
\begin{equation*}
\begin{split}
\tau &= I\alpha\\
TR &= \frac{1}{2}MR^2\alpha\\
\alpha &= \frac{2T}{MR}
\end{split}
\end{equation*}
[/tex]
Now we know that:
[tex]
\begin{equation*}
\begin{split}
a &= R\alpha\\
a &= R(\frac{2T}{MR}) \\
a &= \frac{2T}{M}
\end{split}
\end{equation*}
[/tex]
So...
[tex]
\begin{equation*}
\begin{split}
T &= m(g - \frac{2T}{M})\\
T &= mg(\frac{M}{M + 2m})
\end{split}
\end{equation*}
[/tex]
So am I right in saying that the net torque on the system is
[tex] \tau_{net} = [mg(\frac{M}{M + 2m})]R[/tex]
A 5.40 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 7.00 cm and mass 2.00 kg.
Ok... the first question is:
What is the net torque on the system about the point O?
Here's my take on the situation... the only torque being applied to the system is from the tension T of the string.
T = tension of string
m = mass of counterweight
a = acceleration of counterweight
R = radius of wheel in meters
M = mass of pulley
[tex]T - mg = -ma[/tex]
[tex]T = m(g - a)[/tex]
[tex]
\begin{equation*}
\begin{split}
\sum\tau &= TR \\
&= mR(g - a)
\end{split}
\end{equation*}
[/tex]
Also...
[tex]
\begin{equation*}
\begin{split}
\tau &= I\alpha\\
TR &= \frac{1}{2}MR^2\alpha\\
\alpha &= \frac{2T}{MR}
\end{split}
\end{equation*}
[/tex]
Now we know that:
[tex]
\begin{equation*}
\begin{split}
a &= R\alpha\\
a &= R(\frac{2T}{MR}) \\
a &= \frac{2T}{M}
\end{split}
\end{equation*}
[/tex]
So...
[tex]
\begin{equation*}
\begin{split}
T &= m(g - \frac{2T}{M})\\
T &= mg(\frac{M}{M + 2m})
\end{split}
\end{equation*}
[/tex]
So am I right in saying that the net torque on the system is
[tex] \tau_{net} = [mg(\frac{M}{M + 2m})]R[/tex]