Question

a bullet of mass m and speed v passes completely through a pendulum bob of mass m. the bullet emerges with a speed of v/2. the pendulum bob is suspended by a stiff rod (not a string) of length l and negligible mass. what is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Answer 1

The minimum speed v of a bullet for the pendulum bob to swing through a full vertical circle is determined using the conservation of momentum and energy. It can be calculated with the formula v = √(8gL).

Step-by-step explanation:

Minimum speed v for a pendulum bob to complete a vertical circle can be found by using the principles of conservation of momentum and energy. When the bullet of mass m passes completely through the pendulum bob of the same mass m and emerges with a speed v/2, the pendulum bob will start moving with some speed due to the conservation of momentum. To find the minimum value of v for the pendulum bob to swing through a complete vertical circle, we need to ensure that at the top of the circle, the centripetal force required is provided entirely by the weight of the pendulum bob.

Consequently, it translates to having the gravitational potential energy at the top equal to the initial kinetic energy imparted to the bob by the bullet minus the energy lost by the bullet. That is:

2mgL = 1/2 m (v^2 - (v/2)^2)

After simplifying and solving for v, we'll get the minimum speed v required for the pendulum bob to perform a complete vertical circle:

v = √(8gL)