A brick of mass 1.19 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle has a radius of 1.16 meters. The cable will break if the tension exceeds 45.0 Newtons. What is the maximum speed the brick can have at the bottom of the circle before the cable will break?
To find the maximum speed of the brick at the bottom of the circle before the cable breaks, we need to consider the forces acting on the brick.
At the bottom of the circle, the brick is experiencing two forces: the tension in the cable pointing towards the center of the circle (towards the top) and the gravitational force acting on the brick (towards the bottom).
We can start by calculating the gravitational force acting on the brick using the formula:
Gravitational force = mass × acceleration due to gravity
Since the acceleration due to gravity is approximately 9.8 m/s^2, and the mass of the brick is given as 1.19 kg:
Gravitational force = 1.19 kg × 9.8 m/s^2
Next, we can determine the tension in the cable at the bottom of the circle. At this point, the tension must be at its maximum in order to balance the gravitational force acting on the brick.
Using the equation for centripetal force:
Centripetal force = mass × (velocity^2 / radius)
At the bottom of the circle, the velocity will be at its maximum. Let's assume it is represented by v_max.
Centripetal force = mass × (v_max^2 / radius)
We can equate the gravitational force and the centripetal force, as they must balance out for the brick to remain in circular motion:
Tension = Gravitational force + Centripetal force
Tension = 1.19 kg × 9.8 m/s^2 + 1.19 kg × (v_max^2 / 1.16 m)
Since the tension must not exceed 45.0 Newtons, we can write the equation as:
45.0 N = 1.19 kg × 9.8 m/s^2 + 1.19 kg × (v_max^2 / 1.16 m)
Now, we can solve this equation for v_max.