(1) First compute the linear speed of the mass. If it completes 1 revolution in 0.5 seconds, then the mass traverses a distance of 2π (1.0 m) ≈ 2π m (the circumference of the circular path), so that its speed is
v = (1/0.5 rev/s) • (2π m/rev) = 4π m/s ≈ 12.57 m/s
Then the centripetal acceleration a is
a = v² / (1.0 m) = 16π² m/s² ≈ 160 m/s²
(where r is the path's radius).
(2) By Newton's second law, the tension in the string is T such that
T = (0.50 kg) a = 8π² N ≈ 79 N
(a) The centripetal acceleration of the mass is 158 m/s².
(b) The tension in the string at the given mass is 79 N.
The given parameters;
The angular speed of the attached mass is calculated as follows;
[tex]\omega = \frac{1 \ rev}{0.5 \ s} \times \frac{2\pi \ rad}{1 \ rev} \\\\\omega = 12.57 \ rad/s[/tex]
The centripetal acceleration of the mass is calculated as follows;
[tex]a_c = \omega ^2 r\\\\a_c = (12.57)^2 \times 1\\\\a_c = 158 \ m/s^2[/tex]
The tension in the string is calculated as follows;
[tex]T = ma_c\\\\T = 0.5 \times 158\\\\T = 79 \ N[/tex]
Thus, the tension in the string at the given mass is 79 N.
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