a rod of length 1m and 4 kg mass is fixed at one end and is intially hanging vertically .the free end is now raised until it makes an angle of 60° with the vertical .amount of work required is.
Answers
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5x = 0.25 m so that the vertical lift , h , 0f the centre of mass is h= 0.5–0.25=0.25 m.
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5x = 0.25 m so that the vertical lift , h , 0f the centre of mass is h= 0.5–0.25=0.25 m.W = mgh = [0.5]g[0.25]= 0.125 g N= 9.8/8 n= I.225 Nm [j]
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5x = 0.25 m so that the vertical lift , h , 0f the centre of mass is h= 0.5–0.25=0.25 m.
The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5x = 0.25 m so that the vertical lift , h , 0f the centre of mass is h= 0.5–0.25=0.25 m.The centre of mass is situated at 0.5 m along the rod and is lifted a height h , vertically and x+h =0.5x/0.5 = cos 60 =0.5x = 0.25 m so that the vertical lift , h , 0f the centre of mass is h= 0.5–0.25=0.25 m.W = mgh = [0.5]g[0.25]= 0.125 g N= 9.8/8 n= I.225 Nm [j]
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