At 2 ‘o clock in the afternoon, the hour hand of the clock has a mechanism failure because of which its hour hand (length =0.5m) falls under gravity with its one end being attached to the clock centre, calculate the angular velocity of the hand as it goes past the 6 ‘o clock mark. ()
Answers
Answered by
4
Hello dear,
● Answer- 7.66 rad/s.
● Explaination-
When hour hand is resting at 2 'O clock,
P.E. = mgl.sinθ
Here,θ = angle between 2 'O clock and horizontal = 30°
Therefore,
P.E. = mglsin30
P.E. = mgl/2
When hour hand passes 6 'O clock,
P.E. = -mgl
K.E. = mv^2 / 2
By law of conservation of energy,
Initial energy = Final energy
mgl/2 = -mgl + mv^2/2
v = √(3gl)
Putting values
v = √(3×9.8×0.5)
v = √14.7
v = 3.83 m/s
Angular velocity is
w = v/r
w = 3.83 / 0.5
w = 7.66 rad/s
When hour hand passes 6 'O clock angular velocity will be 7.66 rad/s.
Hope this is useful.
● Answer- 7.66 rad/s.
● Explaination-
When hour hand is resting at 2 'O clock,
P.E. = mgl.sinθ
Here,θ = angle between 2 'O clock and horizontal = 30°
Therefore,
P.E. = mglsin30
P.E. = mgl/2
When hour hand passes 6 'O clock,
P.E. = -mgl
K.E. = mv^2 / 2
By law of conservation of energy,
Initial energy = Final energy
mgl/2 = -mgl + mv^2/2
v = √(3gl)
Putting values
v = √(3×9.8×0.5)
v = √14.7
v = 3.83 m/s
Angular velocity is
w = v/r
w = 3.83 / 0.5
w = 7.66 rad/s
When hour hand passes 6 'O clock angular velocity will be 7.66 rad/s.
Hope this is useful.
Similar questions