A 500 g metal ball is whirled in a vertical circle. What is the centripetal force which draws it toward the center of the circle as it passes through the highest point?
At the top point
F(centr) =mg
4.9 N
To solve this problem, we need to use the concept of centripetal force and consider the forces acting on the metal ball at the highest point of the vertical circle.
Centripetal force is the force that acts towards the center of a circular path, keeping an object in circular motion. In this case, the centripetal force is provided by the tension in the string that holds the metal ball.
At the highest point of the vertical circle, the metal ball experiences two significant forces: the force of gravity and the tension force. The force of gravity acts vertically downward with a magnitude of 500 g, where g represents the acceleration due to gravity (approximately 9.8 m/s^2).
In order to find the centripetal force, we need to determine the net force acting towards the center of the circle. This can be done by considering the forces in the vertical direction. At the highest point, the net force in the vertical direction must be equal to zero to maintain circular motion.
The force of gravity can be resolved into two components: one parallel to the string and one perpendicular to it. The vertical component of the force of gravity (mg) is balanced by the tension force (T), which acts in the upward direction. Therefore, we can write:
T + mg = 0
Since the net force in the vertical direction is zero, the tension force (T) must equal the vertical component of the force of gravity. So, we have:
T = mg
Now, substituting the values:
T = (0.5 kg) x (9.8 m/s^2) = 4.9 N
Therefore, the centripetal force that draws the metal ball towards the center of the circle at the highest point is 4.9 Newtons.