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 :

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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