Question

3) A mass of 0.5 kg is attached to the lower end of a thin cord of which length is 2.1m. The upper end of the cord is fixed. The mass moves in a horizontal circle with constant speed. During the revolution of the mass, the cord makes a fixed angle of 20° with the vertical direction. As a result, the suspending cord traces out a cone. Determine the tension force F in the cord and the period T of the system (time required for one revolution of the mass).

Answer 1

L m=0.5kg Tcoso L = 2.1 m ~mv² Tsina o o = 20° mg Force Sino = L balance gives, Tcoso = mg => 8= L sino = 2.1 Sin 20° T sino = mv2 -0.718 m From 5:22 Ta mg 0.5*9-83 (Here I reters to tension in cord). COSO Cos 20 From m mw²r (i va wo) T sino - ४ 011 from w Dividing o from we get, T sino mg Tcoso m war tano g wer W? 9.81 g stano = 6:217 sad/s: 0718 tan 20° =) | T 27 2T 6.217 1025 sec (Here, I refers to Time period).