I have a question for number 45. I tried something but I am not sure how the gravity would come in and have trouble visualizing it. Would it only be factored for the y-component since it is gravity?
Given:
Mass of Sphere (m) = 2.00 kg
Distance between both the strings (L) = 1.00 m
Considering the given figure and its Free Body Diagram as shown below
Where,
Tension in upper string = T1
Tension in lower string = T2
Length of upper string = L1
Length of lower string = L2
Velocity of ball = v
Radius of horizontal circle = r
a) Considering the FBD as shown in picture,
In x-direction,
T1sin30o + T2sin60o = mv2/r ---(Equation 1)
In y-direction,
T1cos30o + T2cos60o = mg ---(Equation 2)
Also
T1 = T2 ---(Equation 3)
And the given triangle is isosceles triangle (two angle and two sides of a triangle are equal). In this case equal angle = 30o and equal sides are L and L2
Therefore L = L2 = 1.0 m
and r = L2cos30o = 0.866 m ---(Equation 4)
Solving equation 1, 2, 3 and 4 for 'v' we get,
v = (L2cos30o x g)1/2 = 2.91 m/s
For tension to be same in both the wires speed of ball should be 2.91 m/s
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