Question

a 0.13 kg stopper is attched to a 0.93 m length of string. the stopper is swung in.a horizontal circle making one revolution in 1.18 s find the speed of the mass find its centripetal accleration and find the force the string exerts on it

Answer 1

The speed of the stopper is 4.95 m/s, while the centripetal force on the stopper and the tension in the string is 3.42N each.

The mass of the stopper is 0.13 Kg, the length of the string to which it is attached is 0.93 m. The stopper swung and made one revolution in 1.18 seconds, so the time period of revolution of the stopper is 1.18 seconds.

The relation between the time period and the angular speed is,

T = 2π/W

W is angular speed and T is the time period of revolution.

Putting values,

1.18 = 2π/W

W = 5.32 rad/s.

The speed of the stopper can be found as,

v = rW

r is the length of the string,

V = 4.95 m/s.

The centripetal force on the stopper is given by,

F = MV²/r

Putting values,

F = 0.13 x 4.95 x 4.95/0.93

F = 3.42 N.

The tension in the string will be equal to the centripetal force on the stopper, so, the tension is 3.

42 N.

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