A particle of mass 0.20 kg is attached to one end of a light inextensible string of length 50 cm. The particle moves in a horizontal circle with an angular velocity of 5.0 rad s ^ - 1 with the string inclined at theta to the vertical. Find the value of theta

1 answer

The centripetal force acting on the particle is given by:

F = mv^2/r

where m is the mass of the particle, v is its velocity, and r is the radius of the circle.

In this case, the radius is equal to the length of the string, which is 50 cm or 0.5 m. So we have:

F = (0.20 kg)(5.0 rad/s)^2/(0.5 m) = 10 N

This force is acting vertically downwards due to gravity, and is balanced by the tension in the string. We can resolve the tension into two components: one parallel to the string, and one perpendicular to it. The perpendicular component is equal to mg, where g is the acceleration due to gravity.

The parallel component is equal to the centripetal force, which we just calculated. So we have:

t*sin(theta) = mg (perpendicular component)
t*cos(theta) = F (parallel component)

where t is the tension in the string. Dividing the second equation by the first, we get:

tan(theta) = F/mg = 10/(0.20*9.81) ≈ 5.1

Taking the arctan of both sides, we get:

theta ≈ 78.5 degrees