Dear student to calculate angular accelerationT=Iα rFsinθ=Iαθ is angle between radius vector and force vectorhere force is weight and asdius is along length of rod[considering that the disk is attached to the end of the rod at its circumference]I has to be written about centre of rotation = point on rod pivotedI=Inet=Irod about pivot+Idisc about pivot=m2L23+(m1b22+m1(b+L)2)=m2L23+(m1b22+m1(b2+L2+2Lb))=m2L23+(m1b22+m1b2+m1L2+2Lbm1))=m2L23+(m1b2+2m1b2+2m1L2+4Lbm12)=m2L23+(3m1b2+2m1L(L+2b)2)=(2m2L2+9m1b2+6m1L(L+2b)6)=2L(m2+3m1L(L+2b)+9m1b26Fnet=(m1+m2)gr= distance of centre of mass from pivot= m1L2+m2(L+b)(m1+m2)nowrFsinθ=Iα(m1L2+m2(L+b)(m1+m2))(m1+m2)g sinθ2L(m2+3m1L(L+2b)+9m1b26=αm1L2+m2(L+b)g sinθ2L(m2+3m1L(L+2b)+9m1b26=α6(m1L2+m2(L+b)g sinθ)2L(m2+3m1L(L+2b)+9m1b2=α Regards

Q4

Dear student

to calculate angular acceleration

T = I α r F \sin θ = I α θ i s a n g l e b e t w e e n r a d i u s v e c t o r a n d f o r c e v e c t o r h e r e f o r c e i s w e i g h t a n d a s d i u s i s a l o n g l e n g t h o f r o d [c o n s i d e r i n g t h a t t h e d i s k i s a t t a c h e d t o t h e e n d o f t h e r o d a t i t s c i r c u m f e r e n c e] I h a s t o b e w r i t t e n a b o u t c e n t r e o f r o t a t i o n = p o i n t o n r o d p i v o t e d I = I_{n e t} = I_{r o d a b o u t p i v o t} + I_{d i s c a b o u t p i v o t} = \frac{m_{2} L^{2}}{3} + (\frac{m_{1} b^{2}}{2} + m_{1} (b + L)^{2}) = \frac{m_{2} L^{2}}{3} + (\frac{m_{1} b^{2}}{2} + m_{1} (b^{2} + L^{2} + 2 L b)) = \frac{m_{2} L^{2}}{3} + (\frac{m_{1} b^{2}}{2} + m_{1} b^{2} + m_{1} L^{2} + 2 L b m_{1})) = \frac{m_{2} L^{2}}{3} + (\frac{m_{1} b^{2} + 2 m_{1} b^{2} + 2 m_{1} L^{2} + 4 L b m_{1}}{2}) = \frac{m_{2} L^{2}}{3} + (\frac{3 m_{1} b^{2} + 2 m_{1} L (L + 2 b)}{2}) = (\frac{2 m_{2} L^{2} + 9 m_{1} b^{2} + 6 m_{1} L (L + 2 b)}{6}) = \frac{2 L (m_{2} + 3 m_{1} L (L + 2 b) + 9 m_{1} b^{2}}{6} F_{n e t} = (m_{1} + m_{2}) g r = d i s \tan c e o f c e n t r e o f m a s s f r o m p i v o t = \frac{m_{1} \frac{L}{2} + m_{2} (L + b)}{(m_{1} + m_{2})} n o w r F \sin θ = I α \frac{(\frac{m_{1} \frac{L}{2} + m_{2} (L + b)}{(m_{1} + m_{2})}) (m_{1} + m_{2}) g \sin θ}{\frac{2 L (m_{2} + 3 m_{1} L (L + 2 b) + 9 m_{1} b^{2}}{6}} = α \frac{m_{1} \frac{L}{2} + m_{2} (L + b) g \sin θ}{\frac{2 L (m_{2} + 3 m_{1} L (L + 2 b) + 9 m_{1} b^{2}}{6}} = α \frac{6 (m_{1} \frac{L}{2} + m_{2} (L + b) g \sin θ)}{2 L (m_{2} + 3 m_{1} L (L + 2 b) + 9 m_{1} b^{2}} = α

Regards

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