Question 230: Centre of Mass and Rotational Motion

Question

A uniformly thick wheel with moment of inertia
I and radius R is free to rotate about its centre
of mass (see fig). A massless string is wrapped
over its rim and two blocks of masses m
1
and
m
2
(m
1
$ > $ m
2
) are attached to the ends of the
string. The system is released from rest. The
angular speed of the wheel when m
1
descents
by a distance h is :
![](https://imagex.cdn.examgoal.net/7k9k2k5lbfklx/47678b01-fda5-4e56-bdd1-ecb81d9b000c/1453c450-3ae4-11ea-92dd-4b9c3519996f/file-7k9k2k5lbfkly-177w.jpg)

Accepted Answer

By conservation of energy :
Loss in energy = Gain in enrgy
m
1
gh = m
2
gh + ${1 \over 2}$ m
1
V
2
+ ${1 \over 2}$ m
2
V
2
+ ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ (m
1
Ã¢ÂÂ m
2
)gh = ${1 \over 2}$ (m
1
+ m
2
)V
2
+ ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ (m
1
Ã¢ÂÂ m
2
)gh = ${1 \over 2}$ (m
1
+ m
2
)${\left( {\omega R} \right)^2}$ + ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ (m
1
Ã¢ÂÂ m
2
)gh = ${1 \over 2}$ (m
1
+ m
2
)${\left( {\omega R} \right)^2}$ + ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ (m
1
Ã¢ÂÂ m
2
)gh = ${{{\omega ^2}} \over 2}\left[ {\left( {{m_1} + {m_2}} \right){R^2} + I} \right]$
$ \Rightarrow $ $\omega $ = ${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$

Answer

${\left[ {{{2\left( {{m_1} + {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$

Answer

${\left[ {{{{m_1} + {m_2}} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$

Answer

${\left[ {{{\left( {{m_1} - {m_2}} \right)} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$

Answer

${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$

Centre of Mass and Rotational Motion

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