Answer to Question #134796 in Mechanics | Relativity for gee

Question

We have a clock with hour and minute hands. Assume that both vectors point from origin, outward. Length of minute hand is 3.00 cm and hour hand is 6.00 cm. (A) How would you show the resultant vector when the clock's time is 5:30 AM? (B) How would you show the resultant vector in component form when the clock's time is 1:25 PM?

Answer 1

length of minute hand, $l = 3 cm$

length of hour hand, $L = 6 cm$

(a) When it is 5:30 AM, angle between minute and hour hand is $15$ degree.

Resultant of two vectors will be,

$R = \sqrt{l^2 + L^2+2lLcos(15) } = 9.86 cm$

(b) When it is 1:25 PM then angle will be 107.5 degree

then resultant will be,

$R = \sqrt{l^2 + L^2+2lLcos(107.5)} = 5.36 cm$