A 0.50 kg mass is attached to a string 1.0 meter long and moves in a horizontal circle at a rate of 0.5 seconds per revolution. 1 - Find the centripetal acceleration of the mass.

2 - Calculate the tension in the string.

(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;

The centripetal acceleration of the mass is calculated as follows;

The tension in the string is calculated as follows;

Thus, the tension in the string at the given mass is 79 N.

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(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

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find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2