Question:

A small bead of mass 0.2 kg is attached to a string and is being twirled in a horizontal circle with a radius of 0.5 m. The bead completes one revolution in 2 seconds.

1. Calculate the magnitude of the centripetal force acting on the bead.
2. Determine the speed of the bead as it travels in the circular path.
3. If the string is suddenly cut when the bead is at the topmost point of its circular path, describe the motion of the bead after the string is cut.

Answer:

1. To calculate the magnitude of the centripetal force, we will use the formula: F_c = (m * v^2) / r, where F_c is the centripetal force, m is the mass of the bead, v is the linear speed, and r is the radius of the circular path.

Given: Mass of the bead (m) = 0.2 kg Radius of the circular path (r) = 0.5 m Time for one revolution (T) = 2 seconds

First, let's calculate the linear speed of the bead using the formula v = 2πr / T, where π is approximately 3.14.

Plugging in the values, v = (2 * 3.14 * 0.5 m) / 2 s = 3.14 m/s

Now, we can substitute the values of mass (m), linear speed (v), and radius (r) into the centripetal force equation:

F_c = (0.2 kg * (3.14 m/s)^2) / 0.5 m F_c = 0.392 N

Therefore, the magnitude of the centripetal force acting on the bead is 0.392 N.

2. The speed of the bead (v) is already calculated to be 3.14 m/s.

3. If the string is suddenly cut when the bead is at the topmost point of its circular path, the bead would continue moving tangentially along its current direction due to the inertia of motion. It would move in a straight line at a constant speed until acted upon by external forces.

It's important to note that without the centripetal force provided by the string tension, the bead would no longer be in circular motion. Instead, it would move in a path tangent to the circular path at the moment the string was cut.

To summarize, after the string is cut, the bead would move forward in a straight line, tangentially to the circular path it was previously following.