A 0.700 kg object is swung in a circular path and in a vertical plane on a 0.550 m length string. If the angular velocity at the bottom is 12.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?
To find the tension in the string when the object is at the bottom of the circle, we can start by calculating the centripetal acceleration of the object.
The centripetal acceleration (ac) of an object moving in a circle is given by the equation:
ac = ω^2 * r,
where ω is the angular velocity and r is the radius of the circle.
In this case, the angular velocity is given as 12.00 rad/s, and the radius of the circle is equal to the length of the string, which is 0.550 m. Therefore, we have:
ac = (12.00 rad/s)^2 * 0.550 m.
After calculating this, we find that the centripetal acceleration at the bottom of the circle is:
ac = 79.20 m/s^2.
Next, we can calculate the net force acting on the object using Newton's second law of motion:
Fnet = m * ac,
where m is the mass of the object.
The mass of the object is given as 0.700 kg. So we have:
Fnet = 0.700 kg * 79.20 m/s^2.
Calculating this, we get:
Fnet = 55.44 N.
Since the tension in the string provides the centripetal force required to keep the object moving in a circle, the tension in the string at the bottom of the circle is equal to the net force acting on the object:
Tension = Fnet = 55.44 N.
Therefore, the tension in the string when the object is at the bottom of the circle is 55.44 N.