Question 141: Centre of Mass and Rotational Motion

Question

A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and
collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings
as a result of the collision making angle $$	heta $$ before momentarily coming to rest. If the rod has mass
M = 2 kg, and length $$l$$ = 1 m, the value of $$	heta $$ is approximately :
(take g = 10 m/s2)

Accepted Answer

Applying Angular momentum conservation
mv$$l$$ = $$\left( {{{M{l^2}} \over 3} + m{l^2}} ight)$$$$\omega $$
$$ \Rightarrow $$ $$\omega  = {{1 	imes 6 	imes 1} \over {{2 \over 3} + 1}}$$ = $${{18} \over 5}$$
Now, using energy conservation
$${1 \over 2}I{\omega ^2} = \left( {m + M} ight)gh$$
$$ \Rightarrow $$ $${1 \over 2}\left( {{{M{l^2}} \over 3} + m{l^2}} ight){\omega ^2} = \left( {m + M} ight)g{x_{COM}}\left( {1 - \cos 	heta } ight)$$
$$ \Rightarrow $$ $${1 \over 2}\left( {{2 \over 3} + 1} ight){\left( {{{18} \over 5}} ight)^2} = \left( {m + M} ight)g 	imes {{M 	imes {l \over 2} + ml} \over {\left( {M + m} ight)}}\left( {1 - \cos 	heta } ight)$$
$$ \Rightarrow $$ $${1 \over 2} 	imes {5 \over 3} 	imes {{18 	imes 18} \over {25}} = 10 	imes \left( {2 	imes {1 \over 2} + 1} ight)\left( {1 - \cos 	heta } ight)$$
$$ \Rightarrow $$ $${5 \over 6} 	imes {{18 	imes 18} \over {25}} = 20\left( {1 - \cos 	heta } ight)$$
$$ \Rightarrow $$ $$\left( {1 - \cos 	heta } ight) = {{18} \over 5} 	imes {3 \over {20}}$$
$$ \Rightarrow $$ $${\cos 	heta  = 1 - {{27} \over {50}}}$$
$$ \Rightarrow $$ $${\cos 	heta  = {{23} \over {50}}}$$
$$ \Rightarrow $$ $$	heta $$ $$ \simeq $$ 63
o

Answer

63o

Answer

69o

Answer

55o

Answer

49

Centre of Mass and Rotational Motion

Choose the correct answer: