Question-52814

Question Number 52814 by ajfour last updated on 13/Jan/19

Commented by ajfour last updated on 13/Jan/19

Initially rod whose one end is pivoted to centre of disc, is kept vertical and released, Find v(θ). Friction is sufficient say, coefficient is μ.

$${Initially}\:{rod}\:{whose}\:{one}\:{end}\:{is} \\ $$$${pivoted}\:{to}\:{centre}\:{of}\:{disc},\:{is}\:{kept} \\ $$$${vertical}\:{and}\:{released},\:{Find}\:{v}\left(\theta\right). \\ $$$${Friction}\:{is}\:{sufficient}\:{say},\: \\ $$$${coefficient}\:{is}\:\mu. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19

mg×(l/2)(1−cosθ)=(1/2)(((ml^2 )/3))(w_(rod) )^2 +(1/2)mV^2 +(1/2)MV^2 +(1/2)((1/2)MR^2 )(w_(disc) )^2 above eqn loss o P.E=k.E (rotational+liniear )for both disc and rod (((ml^2 )/3))w_(rod) =((1/2)MR^2 )w_(disc) ←conservation of angular momentum (((ml^2 )/3))w_(rod) =(((MR^2 )/2))((V/R)) w_(rod) =((MR^2 ×3)/(2×ml^2 ))×(V/R) w_(rod) =((3M)/(2m))×(((RV)/l^2 )) mg×(l/2)(1−cosθ)=(1/2)(((ml^2 )/3))(((9M^2 )/(4m^2 ))×((R^2 V^2 )/l^4 ))+(1/2)mV^2 +(1/2)MV^2 +(1/2)((1/2)MR^2 )((V^2 /R^2 )) mg×(l/2)(1−cosθ)=(3/8)((M^2 /m))((R^2 /l^2 ))V^2 +(1/2)MV^2 +(1/4)MV^2 =(3/8)((M^2 /m))((R^2 /l^2 ))V^2 +(3/4)MV^2 =(3/4)MV^2 (1+(1/2)×(M/m)×(R^2 /l^2 )) V^2 =((mgl(1−cosθ))/(2×(3/4)M(1+(1/2)×(M/m)×(R^2 /l^2 )))) V^2 =((2mgl(1−cosθ))/(3M(1+((MR^2 )/(2ml^2 ))))) is it ok...

$${mg}×\frac{{l}}{\mathrm{2}}\left(\mathrm{1}−{cos}\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}\right)\left({w}_{{rod}} \right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{MV}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}{MR}^{\mathrm{2}} \right)\left({w}_{{disc}} \right)^{\mathrm{2}} \\ $$$${above}\:{eqn}\:\:{loss}\:{o}\:{P}.{E}={k}.{E}\:\left({rotational}+{liniear}\:\right){for}\:{both}\:{disc}\:{and}\:{rod} \\ $$$$ \\ $$$$\left(\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}\right){w}_{{rod}} =\left(\frac{\mathrm{1}}{\mathrm{2}}{MR}^{\mathrm{2}} \right){w}_{{disc}} \leftarrow{conservation}\:{of}\:{angular}\:{momentum} \\ $$$$\left(\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}\right){w}_{{rod}} =\left(\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{{V}}{{R}}\right) \\ $$$${w}_{{rod}} =\frac{{MR}^{\mathrm{2}} ×\mathrm{3}}{\mathrm{2}×{ml}^{\mathrm{2}} }×\frac{{V}}{{R}} \\ $$$${w}_{{rod}} =\frac{\mathrm{3}{M}}{\mathrm{2}{m}}×\left(\frac{{RV}}{{l}^{\mathrm{2}} }\right) \\ $$$${mg}×\frac{{l}}{\mathrm{2}}\left(\mathrm{1}−{cos}\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}\right)\left(\frac{\mathrm{9}{M}^{\mathrm{2}} }{\mathrm{4}{m}^{\mathrm{2}} }×\frac{{R}^{\mathrm{2}} {V}^{\mathrm{2}} }{{l}^{\mathrm{4}} }\right)+\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{MV}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}{MR}^{\mathrm{2}} \right)\left(\frac{{V}^{\mathrm{2}} }{{R}^{\mathrm{2}} }\right) \\ $$$${mg}×\frac{{l}}{\mathrm{2}}\left(\mathrm{1}−{cos}\theta\right)=\frac{\mathrm{3}}{\mathrm{8}}\left(\frac{{M}^{\mathrm{2}} }{{m}}\right)\left(\frac{{R}^{\mathrm{2}} }{{l}^{\mathrm{2}} }\right){V}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{MV}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{MV}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\left(\frac{{M}^{\mathrm{2}} }{{m}}\right)\left(\frac{{R}^{\mathrm{2}} }{{l}^{\mathrm{2}} }\right){V}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}{MV}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{MV}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{M}}{{m}}×\frac{{R}^{\mathrm{2}} }{{l}^{\mathrm{2}} }\right) \\ $$$${V}^{\mathrm{2}} =\frac{{mgl}\left(\mathrm{1}−{cos}\theta\right)}{\mathrm{2}×\frac{\mathrm{3}}{\mathrm{4}}{M}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{M}}{{m}}×\frac{{R}^{\mathrm{2}} }{{l}^{\mathrm{2}} }\right)} \\ $$$${V}^{\mathrm{2}} =\frac{\mathrm{2}{mgl}\left(\mathrm{1}−{cos}\theta\right)}{\mathrm{3}{M}\left(\mathrm{1}+\frac{{MR}^{\mathrm{2}} }{\mathrm{2}{ml}^{\mathrm{2}} }\right)} \\ $$$${is}\:{it}\:{ok}… \\ $$$$ \\ $$

Commented by ajfour last updated on 13/Jan/19