Answer:
v= 0.667 m/s
Explanation:
M= 0200 Kg, L= 25.0 cm=0.25 cm, θ = 24.0°
To find: V=?
Solution:
according to the situation the formula after deducing the result from the horizontal and vertical components of the force that is
∑(Fx) = Fc= mv²/r = T sinθ ----------- eqn 1
and ∑(Fy) = 0 N= T cosθ - W = T cosθ - mg
T cosθ = mg-------------eqn 2
Dividing eqn 1 by eqn 2 we get
(mv²/r)/mg = T sinθ / T cosθ
⇒ v = square root ( gr tanθ) (r is radius of the circular path that mass will move)
so to find radius r = L sin θ = 0.25 m × sin 24° = 0.102 m
thus v = square root ( 9.81 m/s² × 0.102 m ×tan 24°)
v= 0.667 m/s
The tangential speed of bob to maintain the motion in horizontal circle is 0.667 m/s.
Given data:
The mass of bob is, m = 0.200 kg.
The length of massless string is, L = 25. 0 cm = 0.25 m.
The value of half angle of conical surface is, θ = 24.0 ∘.
The angle made by string with horizontal circle is, θ' = 21°.
As per the given problem, the formula after deducing the result from the horizontal and vertical components of the force is,
∑(Fx) = Fc= mv²/r = T sinθ ........................................................(1)
And,
∑(Fy) = 0 N
T cosθ - W = T cosθ - mg
T cosθ = mg ...........................................................................(2)
taking the ratio of equation 1 and 2 as,
(mv²/r)/mg = T sinθ / T cosθ
here, r is radius of the circular path that mass will move. And its value is,
r = L sin θ
r = 0.25 m × sin 24°
r = 0.102 m
Solving as,
[tex]v =\sqrt{r \times g \times tan \theta}\\\\v =\sqrt{0.102 \times 9.81 \times tan24}\\\\v = 0.667 \;\rm m/s[/tex]
Thus, we can conclude that the tangential speed of bob to maintain the motion in horizontal circle is 0.667 m/s.
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