A small sphere of mass 1 kg is attached to the end of a cord of length 1 m and rotated in a vertical circle about a fixed point O. If the speed at the uppermost point of the trajectory is 1 m/s, the sphere will have a circular trajectory (will move in a circle).
Use g=9.8 m/s2.
I know that this is a true or false question, but I don't know the answer. I will have the answer in just a little bit.
To determine if the sphere will move in a circular trajectory, we need to compare the gravitational force with the tension in the cord at the uppermost point. If the tension is greater than or equal to the gravitational force, the sphere will move in a circular path.
Let's analyze the forces acting on the sphere at the uppermost point of the trajectory:
1. Gravitational Force (Fg):
The gravitational force is given by the equation Fg = mg, where m is the mass of the sphere and g is the acceleration due to gravity. In this case, m = 1 kg and g = 9.8 m/s^2. Therefore, Fg = 1 kg * 9.8 m/s^2 = 9.8 N.
2. Tension Force (T):
At the uppermost point, the net force acting on the sphere must provide the centripetal force to keep it moving in a circular path. The net force is given by the equation Fnet = mv^2/r, where m is the mass of the sphere, v is the speed of the sphere, and r is the radius of the circular path (in this case, the length of the cord). Since the sphere is moving in a horizontal circle, the net force is provided entirely by the tension in the cord.
At the uppermost point, the speed of the sphere is 1 m/s, and the radius of the circular path is 1 m. Therefore, Fnet = 1 kg * (1 m/s)^2 / 1 m = 1 N.
Since the tension in the cord provides the net force, we have T = Fnet = 1 N.
Comparing the gravitational force and the tension force at the uppermost point:
T = 1 N
Fg = 9.8 N
Since T (1 N) is less than Fg (9.8 N), the tension force is not enough to balance the gravitational force, and the sphere will not move in a circular trajectory. It will fall down instead.