Answer:
T = 11.93 N
Explanation:
Newton's second law to the puck in the circular path
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in radial direccion (N)
m : puck mass (kg)
a : radial acceleration of the puck (m/s²)
Data:
m = 0.085 kg
L = 0.72 m = R : radium of the circular path (m)
θ= one revolution = 2Π rad
t= 0.45 s
Angular speed of the puck
ω = θ/t
ω = 1 rev/0.45 s = (2π/0.45) rad/s
ω = 13.96 rad /s
Radial acceleration or centripetal
a = ω²*R
a = (13.96) ²* (0.72)
a = 140.3 m/s²
Magnitude of the tension in the string (T)
We apply the Formula (1)
∑F = m*a
T = (0.085 kg )* (140.3 m/s² )
T = 11.93 N
The magnitude of the tension of the in string is 12.04 N.
The given parameters;
mass of the puck, m = 0.085 kg
length of the string, L = 0.72 m
time for a complete revolution, t = 0.45 s
The magnitude of the tension of the in string is determined by calculating the centripetal force on the circular motion;
[tex]F_c = ma_c = \frac{mv^2}{r}[/tex]
where;
v is the linear velocity (m/s)
r is the radius of the path (m)
The angular speed is calculated as;
[tex]\omega = \frac{1 \ rev}{0.45 \ s} \times \frac{2\pi \ rad}{1 \ rev} = 13.96 \ rad/s[/tex]
The linear velocity is calculated as;
v = ωr
v = 13.96 x 0.72
v = 10.1 m/s
The magnitude of the tension of the in string;
[tex]F_c = \frac{mv^2}{r} = \frac{0.085\times 10.1^2}{0.72} \\\\F_c = 12.04 \ N[/tex]
Thus, the magnitude of the tension of the in string is 12.04 N.
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