A child\'s toy has a 0.104-kg ball attached to two strings, A and B. The strings
ID: 1540568 • Letter: A
Question
A child's toy has a 0.104-kg ball attached to two strings, A and B. The strings are also attached to a stick and the ball swings around the stick along a circular path in a horizontal plane. Both strings are 14.4 cm long and make an angle of 30.0 degree with respect to the horizontal. Draw an FBD for the ball showing the tension forces and the gravitational force. Assume that the ball is to the right of the stick. Find the magnitude of the tension in each string when the ball's angular speed is 3.80 pi rad/s. T_A = N T_B = NExplanation / Answer
On the FBD they drew, A was 210 degrees & B was 150 degrees, I'am not able to draw diagram.
Vertical component = T*sin,
is the angle below the horizontal plane.
The sum of the vertical components of the tension is supporting the weight of the toy as he the toy moves around
the horizontal circle.
Ta*sin a + Tb*sinb = m*g = 0.104*9.81
a = 210
b = 150
m = 0.104 kg
Ta *sin 210 + Tb*sin 150 = 1.02024 ...............(1)
Ta *-0.5 + Tb *0.5 = 1.02024
Divide both sides by 0.5
Ta - Tb = 2.04048
The sum of the 2 horizontal components = total centripetal force
Horizontal component = T * cos
Ta * cosa + Tb*b = mv^2 / r
v = Tangential velocity = angular velocity * radius
v = *r
v^2 = ^2 * r^2
v^2/r = ^2*r
m*v^2/r = m*^2*r
Since each string is hanging at a specific angle below the horizontal plane of the circle in which the toy is moving,
r = Radius = Length * cos
= -30, negative because the strings are 30 below horizontal
m*v^2/r = m*^2*Length*cos
Ta*cosa + Tb*b = m*^2*Length*cos ..................(2)
Ta*cos210 + Tb*150 = 0.104*(3.80*)^2 * 0.144*cos -30
Ta * -0.866 + Tb * -0.866 = 0.104 * (3.80*)^2*0.144*-0.866
Divide both sides by -0.866
Ta + Tb = 0.104*(3.80*)^2 * 0.1247
Ta + Tb = 1.846
Ta - Tb = 2.04048
Add the 2 equations
2*Ta = 3.887
Ta = 1.943 N
1.943 + Tb = 1.846
Tb = -0.097 N
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